A182576 -id:A182576 - cp's OEIS frontend

A182580 Position of first occurrence of n in A182576.

0, 1, 2, 5, 11, 27, 29, 46, 76, 167, 133, 348, 521, 1160, 1364, 3005, 3571, 9348, 9349, 20820, 24476, 64062, 64079, 123651, 167761, 439203, 439204, 1149850, 1149851, 3010348, 3010349, 7881195, 7881196, 20633238, 20633239, 54018520, 54018521, 141422323, 141422324

Offset: 0

Alex Ratushnyak, May 05 2012

nonn

Conjecture: for k > 12, a(2k) = a(2k-1) + 1.

			First occurrence of 4 in A182576(n) has index 11, so a(4)=11.

Cf. A182576.

#include  // GCC // ~880 secs
typedef unsigned long long U64;
U64 fibs[992], first[992], prpr=0, prev=1, current, n, sq;
long long ftop, fpos, terms;
int main(int argc, char **argv)
{
    for (ftop=0; ftop<900; ++ftop) {
      fibs[ftop] = current = prpr+prev;
      if (current0 && fpos>=0; ++terms, sq-=fibs[fpos], --fpos) {
                while (fibs[fpos]>sq && fpos>0) --fpos;
        }
        if (sq>0) { printf("Error!"); exit(1); }
        if (first[terms]==0)
            first[terms]=n, printf("%2llu %llu\n",terms,n);
    }
    for (n=0; n

A182577 Number of ones in Zeckendorf representation of n!

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 3, 5, 6, 9, 8, 11, 11, 11, 16, 17, 17, 18, 23, 23, 28, 31, 33, 27, 33, 29, 40, 37, 42, 42, 41, 44, 47, 44, 53, 56, 57, 50, 64, 55, 59, 68, 63, 72, 70, 61, 69, 85, 80, 83, 87, 97, 98, 101, 87, 91, 100, 102, 114, 108, 116, 109, 117, 117, 113, 124

Offset: 0

Alex Ratushnyak, May 05 2012

nonn

			5! = {1, 0, 0, 1, 0, 1, 0, 0, 1, 0} in the Zeckendorf base.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A007895, A020908, A020910, A025496-A025502, A182535, A182576, A000142.

from math import factorial
def A182577(n):
    m, tlist, s = factorial(n), [1,2], 0
    while tlist[-1]+tlist[-2] <= m:
        tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        if d <= m:
            s += 1
            m -= d
    return s # Chai Wah Wu, Jun 15 2018

A088060 For each Fibonacci number F(m) = A000045(m), m >= 1, look for the smallest Fibonacci number F(k) > F(m) such that F(m) + F(k) is a square. If it exists, append F(k) to the sequence.

Original entry on oeis.org

3, 3, 34, 13, 89, 14930352

Offset: 1

Amarnath Murthy, Sep 21 2003

Conjecture: sequence is finite.
m=1 ==> k=4 or 6, m=2 ==> k=4 or 6, m=3 ==> k=3 or 9, m=4 ==> k=1 or 2 or 7 or 17, m=6 ==> k=1 or 2 or 6, m=7 ==> k=4, m=9 ==> k=3, m=10 ==> k=11, m=11 ==> k=10, m=12 ==> k=36, m=17 ==> k=4, m=36 ==> k=12, for other m there is no solution. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 14 2004
Joe Silverman wrote: "Here's a general finiteness result: fib(n) + c = m^2 has only finitely many solutions, regardless of the value of c. Write fib(n) = a(u^n - u^(-n)), where a=1/sqrt(5) and u=(1+sqrt(5))/2. So generalizing further, can look at the equation au^n + bu^(-n) + c = dm^2, where a,b,c,d are fixed integers (a,b,d nonzero) and u is a unit, or even just an algebraic integer, in some number field.
"Rewrite this as au^(2n) + b + cu^n = d m^2 u^n. Using standard methods, it is not hard to prove that any such equation has only finitely many solutions. One way is to consider the congruence class of n modulo (say) 2. Thus write n = 2*n1+n2. We can assume that n2 is fixed (either 0 or 1). Now let x = u^n1 and y = m be the unknowns and let e=u^n2 be another fixed value.
"Then we need to solve (ae^2)x^4 + b + (ce)x^2 = dx^2y^2. This is the equation of an elliptic curve (albeit a singular one), so has only finitely many solutions in the ring of integers of any field (by a theorem of Siegel). Or one can consider n modulo a higher congruence and then the resulting curve will have genus larger than one, in which case one can quote Faltings' theorem that there are only finitely many points with coordinates in any number field."

			F(10) = 55, F(11) = 89, 55 + 89 = 144 is a square, and the first of the form F(10) + F(k), k > 10. F(10) is the 5th Fibonacci term that has at least one qualifying value, k. Hence a(5) = 89. - _Peter Munn_, Mar 21 2021

Cf. A000045, A094570, A182576.

a(5) and a(6) from Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 14 2004

Name clarified by Peter Munn, Mar 22 2021

A182578 Number of ones in Zeckendorf representation of n^n.

Original entry on oeis.org

1, 1, 2, 3, 3, 6, 3, 10, 13, 12, 16, 15, 20, 24, 20, 30, 25, 31, 26, 33, 33, 31, 34, 42, 49, 49, 53, 55, 56, 55, 58, 64, 64, 67, 73, 78, 70, 76, 77, 75, 89, 83, 92, 90, 106, 99, 100, 99, 107, 116, 107, 115, 125, 125, 122, 119, 127, 137, 127, 138, 155, 156, 153, 160

Offset: 0

Alex Ratushnyak, May 05 2012

nonn

			5^5 = {1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0} in the Zeckendorf base.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A007895, A020908, A020910, A025496-A025502, A182535, A182576, A182577, A000312.

def A182578(n):
    m, tlist, s = n**n, [1,2], 0
    while tlist[-1]+tlist[-2] <= m:
        tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        if d <= m:
            s += 1
            m -= d
    return s # Chai Wah Wu, Jun 14 2018

cp's OEIS Frontend

A182580 Position of first occurrence of n in A182576.

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A182577 Number of ones in Zeckendorf representation of n!

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A088060 For each Fibonacci number F(m) = A000045(m), m >= 1, look for the smallest Fibonacci number F(k) > F(m) such that F(m) + F(k) is a square. If it exists, append F(k) to the sequence.

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A182578 Number of ones in Zeckendorf representation of n^n.

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Author

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Python