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A276318
Least number k such that sigma(Sum_{j=0..n}{k-j}) = sigma(Sum_{j=0..n}{k+j}).
3
1, 17, 89, 79, 321, 49, 23, 20, 139, 87, 26, 48, 41, 56, 75, 88, 38, 81, 49, 134, 196, 78, 68, 114, 155, 65, 552, 111, 80, 349, 103, 104, 142, 192, 110, 342, 123, 359, 395, 249, 83, 349, 101, 74, 481, 292, 219, 1110, 189, 128, 309, 243, 224, 629, 356, 170, 208
OFFSET
0,2
LINKS
FORMULA
Solutions of the equation sigma((n+1)*(2*k-n)/2) = sigma((n+1)*(2*k+n)/2).
EXAMPLE
a(3) = 79 because sigma(76+77+78+79) = sigma(79+80+81+82) = 576.
MAPLE
with(numtheory): P:= proc(q) local k, n; print(1);
for n from 1 to q do for k from n to q do
if sigma((n+1)*(2*k-n)/2)=sigma((n+1)*(2*k+n)/2)
then print(k); break; fi; od; od; end: P(10^9);
MATHEMATICA
Table[k = n; While[DivisorSigma[1, Sum[k - j, {j, 0, n}]] != DivisorSigma[1, Sum[k + j, {j, 0, n}]], k++]; k, {n, 0, 56}] (* Michael De Vlieger, Aug 30 2016 *)
PROG
(PARI) a(n) = {if (n==0, k = 1, k = n); while (sigma((n+1)*(2*k-n)/2) != sigma((n+1)*(2*k+n)/2), k++); k; }
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paolo P. Lava, Aug 30 2016
STATUS
approved