A232538 -id:A232538 - cp's OEIS frontend

A232324 n(n+1)/2 modulo sigma(n).

0, 0, 2, 3, 3, 9, 4, 6, 6, 1, 6, 22, 7, 9, 0, 12, 9, 15, 10, 0, 7, 1, 12, 0, 15, 15, 18, 14, 15, 33, 16, 24, 33, 1, 6, 29, 19, 21, 52, 10, 21, 39, 22, 66, 21, 1, 24, 60, 28, 66, 30, 6, 27, 45, 28, 36, 53, 1, 30, 150, 31, 33, 40, 48, 45, 51, 34, 78, 15, 37, 36

Offset: 1

Jaroslav Krizek, Nov 25 2013

nonn

Also antisigma(n) modulo sigma(n). Antisigma(n) = A024816(n) = the sum of the nondivisors of n that are between 1 and n, sigma(n) = A000203(n) = the sum of the divisors of n.

a(n) = 0 for numbers from A076617, a(n) = 1 for numbers from A232540, a(n) = n for numbers from A232538.

			For n=10, a(10) = antisigma(10) mod sigma(10) = 37 mod 18 = 1.

Jaroslav Krizek, Table of n, a(n) for n = 1..10000

Cf. A000203, A076617, A024816, A232538, A232540.

Table[Mod[n (n + 1)/2, DivisorSigma[1, n]], {n, 100}] (* T. D. Noe, Nov 27 2013 *)

a(n) = n(n+1)/2 mod A000203(n).

A232540 Numbers n such that (n(n+1)/2) modulo sigma(n) = 1.

Original entry on oeis.org

10, 22, 34, 46, 58, 82, 94, 106, 118, 142, 166, 178, 202, 214, 226, 262, 274, 298, 334, 346, 358, 382, 385, 394, 430, 454, 466, 478, 502, 514, 526, 538, 562, 586, 622, 634, 694, 706, 718, 766, 778, 802, 838, 862, 886, 898, 922, 934, 958, 982, 1006, 1018, 1042

Offset: 1

Jaroslav Krizek, Nov 25 2013

nonn

Also numbers n such that antisigma(n) modulo sigma(n) = 1. Antisigma(n) = A024816(n) = the sum of the nondivisors of n that are between 1 and n, sigma(n) = A000203(n) = the sum of the divisors of n.
Numbers n such that A232324(n) = 1.
Number 5950 is only squareful number from first 1400 terms (< 50000) of this sequence.
Conjecture: supersequence of A112774 (semiprimes of the form 6n+4).

			106 is in sequence because antisigma(106) mod sigma(106) = 5509 mod 162 = 1.

Jaroslav Krizek, Table of n, a(n) for n = 1..1000

Cf. A000203, A076617, A024816, A112774, A232324, A232538.

Select[Range[1100],Mod[(#(#+1))/2,DivisorSigma[1,#]]==1&] (* Harvey P. Dale, Sep 08 2017 *)

A378481 Integers k such that A378414(k) == k (mod A066417(k)).

Original entry on oeis.org

33, 77, 153, 372, 1540, 2300, 2692, 2736, 7812, 8721, 12593, 26025, 26481, 27972, 39321, 64009, 104409, 175441, 325180, 335616, 422593, 455625, 564376, 575040, 756460, 800073, 1104521, 2180545, 2304332, 3502665, 3691968, 5130909, 5515121, 9331441, 9546265

Offset: 1

Paolo P. Lava, Nov 28 2024

Also integers k such that A000217(k) == k (mod A066417(k)).

So far, only 33 belongs both to A232538 and A378414.

			Antidivisors of 77 are 2, 3, 5, 9, 14, 17, 22, 31, 51 and their sum is 154.
Then 77*78/2 mod 154 = 3003 mod 154 = 77.

Cf. A000217, A066417, A232538, A378414.

with(numtheory): P:=proc(q) local j,k,n,v; v:=[];
for n from 3 to q do k:=0; j:=n; while j mod 2<>1 do k:=k+1; j:=j/2; od;
if n*(n+1)/2 mod (sigma(2*n+1)+sigma(2*n-1)+sigma(n/2^k)*2^(k+1)-6*n-2)=n
then v:=[op(v),n]; fi; od; op(v); end: P(10^5);

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A232324 n(n+1)/2 modulo sigma(n).

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A232540 Numbers n such that (n(n+1)/2) modulo sigma(n) = 1.

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A378481 Integers k such that A378414(k) == k (mod A066417(k)).

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