A280012 -id:A280012 - cp's OEIS frontend

A351650 Integers m such that digsum(m) divides digsum(m^2) where digsum = sum of digits = A007953.

1, 2, 3, 9, 10, 11, 12, 13, 18, 19, 20, 21, 22, 24, 27, 30, 31, 33, 36, 42, 45, 46, 54, 55, 63, 72, 74, 81, 90, 92, 99, 100, 101, 102, 103, 108, 110, 111, 112, 113, 117, 120, 121, 122, 123, 126, 128, 130, 132, 135, 144, 145, 153, 162, 171, 180, 189, 190, 191, 198

Offset: 1

Bernard Schott, Feb 16 2022

This is a generalization of a problem proposed by French site Diophante in link.
The smallest term k such that the corresponding quotient = n is A280012(n).
The quotient is 1 iff m is in A058369 \ {0}.
If k is in A061909, then digsum(k^2) = digsum(k)^2.
If k is a term, 10*k is also a term.
There are infinitely many m such that both m and m+1 are in the sequence, for example subsequence A002283 \ {0}.
Corresponding quotients are in A351651.

			digit sum of 42 = 4+2 = 6; then 42^2 = 1764, digit sum of 1764 = 1+7+6+4 = 18; as 6 divides 18, 42 is a term.

Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Cf. A004159, A007953, A351651.

Subsequences: A002283, A011557, A052268, A058369, A061909, A093136, A093138, A254066, A280012.

Select[Range[200], Divisible[Total[IntegerDigits[#^2]], Total[IntegerDigits[#]]] &] (* Amiram Eldar, Feb 16 2022 *)

is(n)=sumdigits(n^2)%sumdigits(n) == 0 \\ David A. Corneth, Feb 16 2022

def sd(n): return sum(map(int, str(n)))
def ok(n): return sd(n**2)%sd(n) == 0
print([m for m in range(1, 200) if ok(m)]) # Michael S. Branicky, Feb 16 2022

A004159(a(n)) = A007953(a(n)) * A351651(n).

More terms from David A. Corneth, Feb 16 2022

A351809 a(0) = 32; then, for n >= 1, a(n) is the smallest positive integer k such that pod(k^2)/pod(k) = A002473(n) where pod = product of digits = A007954.

Original entry on oeis.org

32, 1, 2, 3, 15, 381, 25, 61, 12, 27, 16, 41, 28, 23, 336, 13, 1766, 26, 43, 2675, 118, 278, 74, 22, 76, 128, 392, 343, 228, 121, 418, 976, 258, 193, 116, 194, 93, 218, 441, 1231, 112, 63, 219, 984, 136, 4165, 2271, 1894, 183, 615, 434, 22831, 523, 1592, 2435

Offset: 0

Bernard Schott, Feb 24 2022

As pod(m) is a 7-smooth number and pod(m^2) can be 0, all terms of A351808 are in {0} union A002473. See example section for why a(0) = 32.

			pod(32) = 3*2 = 6, pod(32^2) = pod(1024) = 1*0*2*4 = 0, and k = 32 is the smallest positive integer k such that pod(k^2) = 0 while pod(k) <> 0, so a(0) = 32.
A002473(5) = 5; pod(381) = 3*8*1 = 24, pod(381^2) = pod(145161) = 1*4*5*1*6*1 = 120; as 120/24 = 5, and 381 is the smallest positive integer k such that pod(k^2)/pod(k) = 5 then a(5) = 381.
A002473(11) = 12; pod(41)= 4*1 = 4, pod(41^2) = pod(1681) = 1*6*8*1 = 48; as 48/4 = 12 and 41 is the smallest positive integer k such that pod(k^2)/pod(k) = 12, then a(11) = 41.

Cf. A002473, A007954, A280012, A351807, A351808.

sevenSmooths = Select[Range[150], Max[FactorInteger[#][[;; , 1]]] <= 7 &]; pod[n_] := Times @@ IntegerDigits[n]; r[n_] := If[(p = pod[n]) > 0, pod[n^2]/p, -1]; s = Array[r, 3*10^4]; TakeWhile[FirstPosition[s, #] & /@ Join[{0}, sevenSmooths] // Flatten, NumberQ] (* Amiram Eldar, Feb 24 2022 *)

pod(k) = vecprod(digits(k)); \\ A007954
smp(m) = my(k=1); while (!pod(k) || (pod(k^2)/pod(k) != m), k++); k;
isss(n) = (n<11) || (vecmax(factor(n, 7)[, 1])<8); \\ A002473
lista(nn) = apply(smp, select(isss, [0..nn]));
lista(200) \\ Michel Marcus, Feb 24 2022

More terms from Amiram Eldar, Feb 24 2022

A352086 a(n) is the smallest positive integer k such that wt(k^2) / wt(k) = n where wt(k) = A000120(k) is the binary weight of k.

Original entry on oeis.org

1, 21, 2697, 4736533, 14244123157, 4804953862344753

Offset: 1

Bernard Schott, Mar 06 2022

Theorem (proofs in Diophante link):
For any n and any base b, there exists m such that sod_b(m^2) / sod_b(m) = n, where sod_b(m) = sum of digits of m in base b (A280012 for base 10).
a(n) is odd. Proof: a(n) exists. Furthermore, if a(n) is even then wt(a(n)) = wt(a(n)/2) and wt(a(n)^2) = wt((a(n)/2)^2) so then a(n)/2 so that a(n)/2 is a lesser candidate, a contradiction. - David A. Corneth, Mar 06 2022

			We have 21_10 = 10101_2, so wt(21) = 3 ones; then 21^2 = 441_10 = 110111001_2, so wt(21^2) = 6 ones; as 6/3 = 2 and 21 is the smallest integer k such that wt(k^2) / wt(k) = 2, hence a(2) = 21.

Diophante, A1730 - Des chiffres à sommer pour un entier (in French).
Wojciech Muła, Nathan Kurz and Daniel Lemire, Faster Population Counts using AVX2 Instructions, arXiv:1611.07612 [cs.DS], Sep 05 2018.

Cf. A000120, A077436, A083567, A159918, A280012, A352084, A352085.

r[n_] := Total[IntegerDigits[n^2, 2]]/Total[IntegerDigits[n, 2]]; seq[max_, nmax_] := Module[{s = Table[0, {max}], c = 0, n = 1, i}, While[c < max && n < nmax, i = r[n]; If[IntegerQ[i] && s[[i]] == 0, c++; s[[i]] = n]; n+=2]; TakeWhile[s, # > 0 &]]; seq[4, 5*10^6] (* Amiram Eldar, Mar 06 2022 *)

from gmpy2 import popcount
aDict=dict()
for k in range(1, 10**11, 2):
    if popcount(k*k)%popcount(k)==0:
        n=popcount(k*k)//popcount(k)
        if not n in aDict:
            print(n, k); aDict[n]=k # Martin Ehrenstein, Mar 16 2022

a(n) > 2^(n^2/2) for n > 1. - Charles R Greathouse IV, Mar 16 2022

a(3)-a(5) from David A. Corneth, Mar 06 2022

a(6) -- using the Muła et al. Faster Population Counts algorithm -- from Martin Ehrenstein, Mar 15 2022

A280006 Number of possible connected steric graphs with n nodes of degree less than 5, excluding the ones with all nodes of degree 4 and trees.

Original entry on oeis.org

0, 0, 1, 6, 39, 399

Offset: 1

Natan Arie Consigli, Jan 02 2017

Also number of hypothetical neutrally charged hydrocarbons with n carbon atoms containing a cyclic set of atoms (cycloalkanes and alkyl-cycloalkanes) satisfying the octet rule. This includes all isomers except conformers.
Steric (three-dimensional) graphs can be represented as two-dimensional graphs provided that stereoisomers are taken into account. This means acting the group S_4 on the stereocenters (nodes with 4 different substituents) of each 2D graph and counting all of the resulting non-symmetric molecules.
The terms of this sequence (and related ones) were found by drawing and counting the hydrocarbons in the PDFs below. If there are any missing molecules you can contact me. I'll add them in the PDF and update the data section.
Many of those "theoretical" molecules are to be discovered and some may not exist due to their strained geometries.

Natan Arie' Consigli, Illustration of a(5) = 39: molecules with 5 carbons
Natan Arie' Consigli, Illustration of a(6) = 399: molecules with 6 carbons

Cf. A000628 (counts acyclic alkanes), A000639 (counts alkyl-benzenes).

a(n) = A280012(n) - A000628(n).

A351651 a(n) is the quotient obtained when digsum(m^2) is divided by digsum(m), with digsum = sum of digits = A007953 and m = A351650(n).

Original entry on oeis.org

1, 2, 3, 1, 1, 2, 3, 4, 1, 1, 2, 3, 4, 3, 2, 3, 4, 3, 2, 3, 1, 1, 2, 1, 3, 2, 2, 2, 1, 2, 1, 1, 2, 3, 4, 2, 2, 3, 4, 5, 3, 3, 4, 5, 3, 3, 2, 4, 3, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 2, 3, 4, 3, 3, 2, 3, 4, 5, 3, 2, 4, 5, 2, 2, 3, 3, 3, 3, 2, 2, 2, 3, 2, 1, 3, 4, 3, 4, 5

Offset: 1

Bernard Schott, Feb 16 2022

All positive integers are terms of this sequence (see A280012).
a(n) = 1 iff m = A351650(n) is a term of A058369 \ {0}.
a(n) = digsum(n) if m = A351650(n) is a term of A061909 \ {0}.

			A351650(8) = 13, then digsum(13) = 1+3 = 4 while digsum(13^2) = digsum(169) = 1+6+9 = 16; hence, a(8) = 16/4 = 4.

Diophante, A1730 - Des chiffres à sommer pour un entier (in French).

Cf. A002283, A004159, A007953, A058369, A061909, A254066, A280012, A351650.

Select[Total[IntegerDigits[#^2]]/Total[IntegerDigits[#]]& /@ Range[300], IntegerQ] (* Amiram Eldar, Feb 16 2022 *)

lista(nn) = {my(list = List(), q); for (n=1, nn, if (denominator(q=sumdigits(n^2)/sumdigits(n))==1, listput(list, q));); Vec(list);} \\ Michel Marcus, Feb 16 2022

a(n) = A004159(A351650(n)) / A007953(A351650(n)).

More terms from Michel Marcus, Feb 16 2022

cp's OEIS Frontend

A351650 Integers m such that digsum(m) divides digsum(m^2) where digsum = sum of digits = A007953.

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A351809 a(0) = 32; then, for n >= 1, a(n) is the smallest positive integer k such that pod(k^2)/pod(k) = A002473(n) where pod = product of digits = A007954.

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A352086 a(n) is the smallest positive integer k such that wt(k^2) / wt(k) = n where wt(k) = A000120(k) is the binary weight of k.

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A280006 Number of possible connected steric graphs with n nodes of degree less than 5, excluding the ones with all nodes of degree 4 and trees.

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A351651 a(n) is the quotient obtained when digsum(m^2) is divided by digsum(m), with digsum = sum of digits = A007953 and m = A351650(n).

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