SuperSums & SuperProducts
The hereunder mail was sent recently to Math-Fun.
Hello Math-Fun,
say we call a(n) and a(n+1) two successive terms of S, and SuperSum the sum of the digits of a(n) and a(n+1) — the SuperSum of (710;2023) is equal to 7+1+0+2+0+2+3=15.
We want to build now the lexicographically earliest sequence of distinct terms > 0 such that the SuperSum of [a(n); a(n+1)] divides exactly a(n+2).
If one starts S with 1 and 2, I guess we have:
S = 1,2,3,5,8,13,12,7,10,16,24,26,14,39,17,20,30,15,27,45,36,…
Question:
Is S (not in the OEIS) a permutation of the whole numbers?
(I would say yes)
Hans H. was quick to correct the above sequence and commented:
> EA: "S = 1, 2, 3, 5, 8, 13, 12, 7, 10, 16, 24, 26, 14, 39, 17, 20, 30, 15, 27, 45, 36, ..."
HH >Replace 27 with 9 and 36 with 18.
> EA: "Is S (not in the OEIS) a permutation of the whole numbers? (I would say yes.)"
HH > I would say no. How likely are smallish numbers (like 4) to appear as the emerging terms get larger?
> EA: "... I was wondering about the possibility that two successive terms of S could have a SuperSum of, say, 4 (first term = 100001 for instance and second term = 20000)... Dunno if this is possible..."
HH > We don't know if 100001 is ever going to appear. We don't know of 20000 is ever going to appear. But even if both should happen to appear, it strikes me as unlikely that they would appear right next to each other.
Then arrived Maximilian's mail:
MH > So the sequence is a(+1) = least unused multiple of sod(a(n))+sod(a(n-1)).
You can paste this into http://pari.math.u-bordeaux.fr/gp.html :
{a=List(u=[1,2]); for(n=2,99, s=sumdigits(a[n-1])+sumdigits(a[n]);
forstep(k=s,oo,s, set search (u,k)&& next; listput(a,k); u=setunion(u,[k]);
break));a}
It gives:
%1 = List([1, 2, 3, 5, 8, 13, 12, 7, 10, 16, 24, 26, 14, 39, 17, 20, 30, 15, 9, 45, 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198, 27, 81, 216, 234, 252, 270, 288, 135, 189, 243, 297, 324, 351, 306, 342, 360, 378, 405, 432, 396, 459, 468, 504, 486, 513, 540, 414, 450, 522, 558, 567, 576, 612, 594, 621, 648, 675, 684, 720, 702, 630, 666, 729, 756, 792, 828, 864, 900, 783, 810, 837, 891, 936, 972, 1008, 918, 945, 1044, 999, 1080, 1116, 738, 1026, 1053, 774, 1107, 1134, 846, 1161])
Update: (once again the mail remained unsent in "drafts"...)
> I agree with Hans, there is very low probability that the small terms that don't occur "early" will occur later.
> In order for a(n) = 100001 to be followed by 20000, the latter would have to be the least unused multiple of digitsum(a(n-1)) + 2 ... and in turn, 100001 = 11 * 9091 would have to be the smallest unused multiple of the digit sum of the two preceding terms... highly improbable ! > FWIW, the numbers 4, 6, 11 and then 19 don't occur among the first 10^4 terms;
a(..., 10001) = (184905, 184950, 171396, 171504, 184995, 171558, 45927, 171612, 185040, 153756), where 184995 and 185040 are the largest terms to occur so far.
If accepted, this will be https://oeis.org/A364120
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Update #1 (of this page, by EA, after this post by Hans on Math-Fun):
MH:
> You can paste this into http://pari.math.u-bordeaux.fr/gp.html :
> {a=List(u=[1,2]); for(n=2,99, s=sumdigits(a[n-1])+sumdigits(a[n]);
> forstep(k=s,oo,s, set search (u,k)&& next; listput(a,k); u=setunion(u,[k]);
> break));a}
HH:
... Thank you, Max. The web app had no problem calculating 100000 terms. Here's a Mathematica ListPlot of them:
«Seven threads have now been colourized and identified»
The smallest numbers that haven't appeared thus far are: 4, 6, 11, 19, 21, 22, 23, 25, 28, 29, 31, 32, 33, 34, 35, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 127, 128, 129, 130, ...
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Update (of the above list with more terms, by Hans H., Friday 14th,2023)
>As per the previously noted "starting with a(19), every sum-of-digits is a multiple of 9", I can now refine my "there will be more" with:
>231 multiples of 9 (less than 10000) have yet to appear: 117, 153, 171, 207, 261, 279, 333, 369, 387, 423, 477, 531, 549, 603, 639, 657, 711, 747, 801, 873, 909, 927, 963, 981, 1017, 1143, 1179, 1233, 1251, 1341, 1359, 1413, 1467, 1503, 1521, 1557, 1611, 1629, 1719, 1737, 1773, 1791, 1899, 1989, 2007, 2043, 2061, 2097, 2151, 2169, 2223, 2259, 2313, 2367, 2421, 2439, 2493, 2529, 2547, 2601, 2637, 2691, 2763, 2799, 2817, 2853, 2907, 2979, 3033, 3123, 3141, 3177, 3231, 3249, 3303, 3357, 3393, 3411, 3447, 3501, 3519, 3573, 3609, 3627, 3681, 3771, 3789, 3879, 3897, 3933, 3951, 3987, 4041, 4113, 4149, 4167, 4203, 4311, 4329, 4383, 4419, 4437, 4491, 4527, 4581, 4689, 4707, 4743, 4761, 4797, 4869, 4923, 4959, 5013, 5031, 5067, 5121, 5139, 5193, 5283, 5301, 5337, 5391, 5409, 5463, 5499, 5517, 5553, 5571, 5661, 5679, 5769, 5787, 5823, 5877, 5931, 5949, 6003, 6057, 6093, 6147, 6201, 6219, 6273, 6309, 6327, 6381, 6417, 6471, 6543, 6579, 6597, 6651, 6687, 6759, 6813, 6849, 6903, 6921, 6957, 7011, 7083, 7137, 7173, 7191, 7281, 7299, 7353, 7389, 7407, 7443, 7461, 7551, 7569, 7659, 7677, 7713, 7731, 7767, 7839, 7893, 7929, 7947, 7983, 8037, 8091, 8109, 8163, 8199, 8271, 8307, 8361, 8433, 8469, 8487, 8523, 8541, 8577, 8649, 8703, 8739, 8793, 8847, 8901, 8919, 8973, 9027, 9063, 9081, 9117, 9171, 9189, 9243, 9279, 9297, 9333, 9351, 9441, 9459, 9549, 9567, 9621, 9657, 9711, 9729, 9783, 9819, 9837, 9873, 9927, 9981.
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Update #2 (same day, a couple of minutes later) with this communication (in private) by Hans and his link to a wonderful table he computed:
> The b-file of your sequence that previously had a sum-of-digits column added now has an "ss" column as well < http://chesswanks.com/num/SuperSumDividing.txt > so that you can read off the "super sum" for each entry. For example, in my very capable text app, if I search for occurrences of ss = 99 ("99\r" where \r is my hard line return), it gives me the 944 instances. Because the largest of these are < 262000, they are part of the black points at the bottom of my graph. If I search for occurrences of ss = 108, I get 16 instances, some of which are fairly large but don't show in the graph, presumably because of their small point size and density.
[It occurs to me that they might also be lost in the ss = 54 thread, in the same manner that the ss = 90 and ss = 45 threads occupy mostly the same space.]
ÉA: the graph is still here: http://chesswanks.com/num/SuperSumDividing.png
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MH:
>Very nice!
Do you have more explanations on your blog or just the plot?
It would be interesting to describe/explain the different "rays" (not really rays, rather sub-graphs, I hope you know what I mean...?)
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Many thanks, Hans and Maximilian – wonderful graph indeed!
And now for something not completely different
I looked for other sequences that could be inspired by the concept of SuperSum. And rediscovered this 3-year old submission by David James Sycamore:
«Lexicographically earliest sequence of distinct positive integers such that the sum of digits of any consecutive pair of terms divides their consecutive concatenation.»
S = 1, 2, 4, 5, 10, 8, 20, 7, 11, 16, 30, 6, 3, 12, ... [https://oeis.org/A335315]
Great minds, etc.
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What about the pair of successive terms (A;B) whose SuperSum divides A+B?
In other words, we want that the sum of the digits present in the pair (A;B) divides exactly the sum A+B. Maximilian gets T with the hereunder PARI code:
{first(N, U=[], a)=vector(N,n, a=if(n>1, U=setunion(U,[a]); while(#U>1&&U[1]+1==U[2],U=U[^1]);
my(s=sumdigits(a), k=U[1]); while(setsearch(U,k++) || (k+a)%(sumdigits(k)+s),); k, 1))}
T = 1, 2, 3, 4, 5, 6, 7, 8, 9, 27, 21, 15, 12, 24, 30, 10, 11, 13, 14, 22, 20, 16, 32, 28, 26, 34, 38, 46, 44, 40, 23, 25, 29, 31, 17, 19, 35, 37, 47, 43, 41, 49, 59, 76, 50, 55, 53, 52, 56, 70, 60, ...
Check
(...)
a(8) = 8 and a(9) = 9
SS: 8 + 9 = 17
Sum 8 + 9 = 17 ... yes, 17/17 = 1 is a whole number. Then:
a(9) = 9 and a(10) = 27
SS: 9 + 2 + 7 = 18
Sum 9 + 27 = 36 ... yes, 36/18 = 2 is a whole number. Then:
a(10) = 27 and a(11) = 21
SS: 2 + 7 + 2 + 1 = 12
Sum 27 + 21 = 48 ... and yes, 48/12 = 4 is a whole number. Then:
a(11) = 45 and a(12) = 15
SS: 4 + 5 + 1 + 5 = 15
Sum 45 + 15 = 60 ... and yes, 60/15 = 4 is a whole number. Etc.
If accepted, this will be https://oeis.org/A364187
What about SuperProducts?
A SuperProduct involves a pair of successive terms whose digits are successively multiplied. The SuperProduct of the pair (711, 2023) is zero. The SP of the pair (912,1951) is 9*1*2*1*9*5*1 = 810.
Let's play with SuperSums dividing SuperProducts.
We are looking for the lexicographically earliest sequence U of distinct positive integers such that the sum of the digits present in a(n) and a(n+1) divides exactly the product of the same digits. Maximilian was of great help here too (PARI code hereunder and extension of U):
{first(N, U=[], P(x)=vecprod(digits(x)), a)=vector(N,n, a=if(n>1, U=setunion(U,[a]);
while(#U>1&&U[1]+1==U[2],U=U[^1]); my(s=sumdigits(a), p=P(a), k=U[1]);
while(setsearch(U,k++) || P(k)*p%(sumdigits(k)+s),); k, 1))}
U = 1, 10, 2, 13, 8, 19, 5, 14, 12, 3, 6, 17, 20, 4, 15, 9, 18, 16, 7, 25, 21, 30, 11, 24, 28, 33, 40, 22, 26, 31, 23, 27, 32, 34, 29, 37, 38, 41, 43, 35, 44, 50, 36, 45, 47, 46, 42, 39, 48, 53, 52, 49, 56, 54, 60, 51, 63, 57, 58, 61, 65, 59, 64, 55, 68, 70, 62, 67, 69, 66, 72, 80, 71, 76, 74, 73, 83, 79, 84, 90, 75, 85, 78, 89, 87, 93, 96, 81, 99, 100, 77, 86, 82, 95, 88, 97, 92, 94, 101, 91, ...
Check
a(1) = 1 and a(2) = 10
SP: 1 * 1 * 0 = 0
SS: 1 + 1 + 0 = 2 ... and yes, 0/2 = 0 is a whole number. Then:
a(2) = 10 and a(3) = 2
SP: 1 * 0 * 2 = 0
SS: 1 + 0 + 2 = 3 ... and yes, 0/3 = 0 is a whole number. Then:
a(3) = 2 and a(4) = 13
SP: 2 * 1 * 3 = 6
SS: 2 + 1 + 3 = 6 ... and yes, 6/6 = 1 is a whole number. Then:
a(4) = 13 and a(5) = 8
SP: 1 * 3 * 8 = 24
SS: 1 + 3 + 8 = 12 ... and yes, 24/12 = 2 is a whole number. Then:
a(5) = 8 and a(6) = 19
SP: 8 * 1 * 9 = 72
SS: 8 + 1 + 9 = 18 ... and yes, 72/18 = 4 is a whole number. Then:
a(6) = 19 and a(7) = 5
SP: 1 * 9 * 5 = 45
SS: 1 + 9 + 5 = 15 ... and yes, 45/15 = 3 is a whole number. Etc.
Merci beaucoup again, Maximilian!
If accepted, this will be https://oeis.org/A364188
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Anyone wanting to extend the variants V and W and X hereunder?
______________________________________________
Update #2 22:49 (Brussels time)
Three variants (I am almost 100% sure they are not the lexicographically earliest ones, sorry).
Variant V:
«Lexicographically earliest sequence V such that the sum of the digits of a(n) and a(n+1) divides exactly a(n+1).»
V = 1, 10, 12, 20, 14, 26, 38, 32, 33, 24, 40, 21, 30, 36, 11, 80, ...
Check
a(1) = 1 and a(2) = 10
SS: 1 + 1 + 0 = 2 ... yes, 10/2 = 5 is a whole number. Then:
a(2) = 10 and a(3) = 12
SS: 1 + 0 + 1 + 2 = 4 ... yes, 12/4 = 3 is a whole number. Then:
a(3) = 12 and a(4) = 20
SS: 1 + 2 + 2 + 0 = 5 ... and yes, 20/5 = 4 is a whole number. Then:
a(4) = 20 and a(5) = 14
SS: 2 + 0 + 1 + 4 = 7 ... and yes, 14/7 = 2 is a whole number. Then:
a(5) = 14 and a(6) = 26
SS: 1 + 4 + 2 + 6 = 13 ... and yes, 26/13 = 2 is a whole number. Etc.
Variant W:
«Lexicographically earliest sequence W such that the sum of the digits of the pair [a(n); a(n+1)] divides exactly the largest integer of the pair [a(n);a(n+1)].»
W = 1, 10, 4, 21, 13, 9, 11, 14, 2, 20, 3, 12, 30, 7, 22, 28, 34, 19, 18, ...
Check
a(1) = 1 and a(2) = 10
SS: 1 + 1 + 0 = 2 ... yes, 10/2 = 5 is a whole number. Then:
a(2) = 10 and a(3) = 4
SS: 1 + 0 + 4 = 5 ... yes, 10/5 = 2 is a whole number. Then:
a(3) = 4 and a(4) = 21
SS: 4 + 2 + 1 = 7 ... and yes, 21/7 = 3 is a whole number. Then:
a(4) = 21 and a(5) = 13
SS: 2 + 1 + 1 + 3 = 7 ... and yes, 21/7 = 3 is a whole number. Then:
a(5) = 13 and a(6) = 9
SS: 1 + 3 + 9 = 13 ... and yes, 13/13 = 1 is a whole number. Etc.
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Update #3 (next day): the X Variant
A last sequence that plays with the idea of SuperSum (this name is ridiculous! I think InfraSum and InfraProduct are just a bit better).
«The sum of the digits used to write a(n) and a(n+1) is a palindrome. This is the lexicographically earliest sequence of distinct positive terms having this property»:
X = 1, 2, 3, 4, 5, 6, 10, 7, 11, 9, 20, 12, 8, 21, 13, 14, 15, 23, 22, 16, 31, 25, 40, 30, 17, 59, 26, 68, 35, 77, 44, 86, 53, 95, 62, 100, 19, 39, 28, 48, 37, 57, 46, 66, 55, 75, 64, 84, 73, 93, 82, 129, 91, 138, 109, 147, 118, 156, 127, 165, 136, 174, 145, 183, 154, 192, 163, 219, ...
Check
a(1) = 1 and a(2) = 2
SS: 1 + 2 = 3 and 3 is a palindrome. Then:
a(2) = 2 and a(3) = 3
SS: 2 + 3 = 5 and 5 is a palindrome. Then:
(...)
a(6) = 6 and a(7) = 10
SS: 6 + 1 + 0 = 7 and 7 is a palindrome. Then:
a(7) = 10 and a(8) = 7
SS: 1 + 0 + 7 = 8 and 8 is a palindrome. Then:
a(8) = 7 and a(9) = 11
SS: 7 + 1 + 1 = 9 and 9 is a palindrome. Then:
a(9) = 11 and a(10) = 9
SS: 1 + 1 + 9 = 11 and 11 is a palindrome. Etc.
If accepted, this will be https://oeis.org/A364189
(many thanks to Giorgos Kalogeropoulos)
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The forgotten Y variant:
«The sum of the digits present in a(n) and a(n+1) divides the product [a(n)*a(n+1)]. This is the lexicographically earliest sequence of distinct positive terms with this property»:
Y = 1, 10, 4, 15, 3, 6, 12, 9, 11, 14, 2, 20, 8, 26, 5, 32, 21, 13, 18, 16, 7, 22, 25, 30, 24, 28, 31, 40, 33, 23, 50, 41, 95, 44, 38, 17, 27, 19, 34, 37, 66, 42, 35, 48, 39, 60, 36, 45, 51, 29, 78, 47, 138, 55, 64, 105, 52, 57, 43, 70, 58, 46, 49, 75, 63, 54, 72, 65, 96, 81, 84, 79, 115, 76, 114, 67, 160, 85, 103, 100, ... ...
Check:
digitsum a(1) + digitsum a(2) = 1 + 1 + 0 = 2 and 2 divides 1 * 10 = 10 (result = 5);
digitsum a(2) + digitsum a(3) = 1 + 0 + 4 = 5 and 5 divides 10 * 4 = 40 (result = 8);
digitsum a(3) + digitsum a(4) = 4 + 1 + 5 = 10 and 10 divides 4 * 15 = 60 (result = 6);
digitsum a(4) + digitsum a(5) = 1 + 5 + 3 = 9 and 9 divides 15 * 3 = 45 (result = 5);
digitsum a(5) + digitsum a(6) = 3 + 6 = 9 and 9 divides 3 * 6 = 18 (result = 2); etc.If accepted this will be https://oeis.org/A364190.
(many thanks again to Giorgos Kalogeropoulos)
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Update #4:
The Z variant was computed by hand this morning (the usual caveat to the reader):
«The sum of the digits present in a(n) and a(n+1) divides exactly a(n). This is the lexicographically earliest infinite sequence of distinct positive terms with this property»:
Z = 10, 13, 18, 27, 99, 69, 17, 36, 12, 21, 22, 16, 45, 15, 54, 63, 39, 999, 19, 72, 30, 25, 189, 81, 198, 31, 1899, 499999999999999999999, ...
If we want the sequence to be infinite, we cannot extend it with terms < 10. After a(28) = 499999999999999999999 the terms become exponentially huge and impossible to present here.
Check
digitsum a(1) + digitsum a(2) = 1 + 0 + 1 + 3 = 5 and 5 divides exactly a(1) = 10;
digitsum a(2) + digitsum a(3) = 1 + 3 + 1 + 8 = 13 and 13 divides exactly a(2) = 13;
digitsum a(3) + digitsum a(4) = 1 + 8 + 2 + 7 = 18 and 18 divides exactly a(3) = 18;
digitsum a(4) + digitsum a(5) = 2 + 7 + 9 + 9 = 27 and 27 divides exactly a(4) = 27;
digitsum a(5) + digitsum a(6) = 9 + 9 + 6 + 9 = 33 and 33 divides exactly a(5) = 99; etc.
If accepted by the OEIS, this will be https://oeis.org/A363079.
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Update #5:
Yesterday, after dinner (Brussels time), I bumped into the hereunder message. I hesitated a bit this morning, then decided to unsubscribe from Math-Fun (indeed this affair sadly reminded me of my exclusion from the SeqFan mailing list a couple of years ago – it started exactly with a message like that). Now it's me who takes the lead because... who needs floods today (excepts non-digital lands)? I will try to find another platform/forum that might accept my posts (at least for a few months!)
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