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A224928 Numbers of pairs {x, y} such that x <= y and triangular(x) + triangular(y) = 2^n. 2
1, 1, 1, 0, 2, 0, 1, 0, 3, 0, 2, 0, 4, 0, 1, 0, 8, 0, 2, 0, 4, 0, 4, 0, 8, 0, 2, 0, 24, 0, 2, 0, 8, 0, 8, 0, 8, 0, 2, 0, 32, 0, 4, 0, 16, 0, 4, 0, 32, 0, 4, 0, 32, 0, 4, 0, 4, 0, 8, 0, 16, 0, 2, 0, 32, 0, 6, 0, 48, 0, 16, 0, 16, 0, 8, 0, 384, 0, 4, 0, 16, 0, 16, 0, 16, 0, 8, 0, 768, 0, 2, 0, 8, 0, 4, 0, 32, 0, 32, 0, 256 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,5
COMMENTS
Conjectures:
1. a(n) = 0 for odd n > 1.
2. a(n) is even for even n > 14.
LINKS
FORMULA
a(n) = A052343(2^n).
EXAMPLE
2^1 = 1 + 1, the only representation of 2 as a sum of two triangular numbers, so a(1)=1.
2^4 = 16 = 1+15 = 6+10, two representations, so a(4) = 2.
2^8 = 256 = 3+253 = 66+190 = 120+136, so a(8) = 3.
2^12 = 4096 = 1+4095 = 91+4005 = 1540+2556 = 2016+2080, so a(12) = 4.
PROG
(C)
#include <stdio.h>
#include <math.h>
typedef unsigned long long U64;
U64 isTriangular(U64 a) { // ! Must be a <= (1<<63)
U64 s = sqrt(a*2);
if (a>=(1ULL<<63)) {
if (a==(1ULL<<63)) return 0;
printf("Error: a = %llu\n", a), exit(1);
}
return (s*(s+1)/2 == a);
}
int main() {
U64 c, n, x, tx;
for (n = 1; n; n+=n) {
for (c = x = tx = 0; tx*2 <= n; ++x, tx+=x)
if (isTriangular(n - tx))
++c; //, printf("(%llu+%llu) ", tx, n-tx);
printf("%llu, ", c);
}
return 0;
}
(PARI)
A008441(n) = if(!n, n, sumdiv(4*n + 1, d, (d%4==1) - (d%4==3)));
A052343(n) = if(!n, 1, my(u=A008441(n)); ((u\2)+(u%2)));
A224928(n) = A052343(2^n); \\ Antti Karttunen, May 24 2021
CROSSREFS
Sequence in context: A120616 A108044 A104477 * A308066 A052173 A177825
KEYWORD
nonn
AUTHOR
Alex Ratushnyak, May 08 2013
EXTENSIONS
More terms from Antti Karttunen, May 24 2021
STATUS
approved

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Last modified July 18 21:02 EDT 2024. Contains 374388 sequences. (Running on oeis4.)