A367420 -id:A367420 - cp's OEIS frontend

A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)(n + 2k)/2, or a(n) = -1 if no such k exists.

0, 1, 3, 34, 55, 21, 21, -1, 144, 55, 55, -1, 377, 987, 6765, 2584, 2584, -1, 2584, 610, 987, 6765, 46368, -1, 75025, 377, 14930352, -1, 317811, 6765, 832040, 14930352, 6765, -1, 102334155, -1, 4181, -1, 317811, -1, 6765, 987, 701408733, -1, 1548008755920, -1, 2178309, -1, 225851433717

Offset: 0

Ctibor O. Zizka, Jul 17 2025

sign

n even: F_i exists for all even n. Proof: For n even we have (n*(n + 1)/2) mod (n + 1) = 0. The residue 0 is among the residues of the Pisano period of (Fibonacci sequence mod (n + 1)) for all  n.
n odd: F_i exists only if the residue (n*(n + 1)/2) mod (n + 1) = (n + 1)/2 is among the residues of the Pisano period of (Fibonacci sequence mod (n + 1)).
Computational result: it looks like a(2*A367420(n) - 1) = -1, checked for n from [1, 2000].
The lower bound for nonnegative a(n) is n*(n + 1)/2.
Empirical observation: for a(n) = F_i, i = t*A001177(n + 1), t is from {1/3, 1/2, 2/3, 1, 3/2, 2}.

			For n = 3: (3 + 1)*(3 + 2*k)/2 = 6 + 4*k is a Fibonacci number at first for k = 7. Thus a(3) = 34.
For n = 7: (7 + 1)*(7 + 2*k)/2 = 28 + 8*k. (Fib_i congruent to 4 mod 8) has no solution because the Fibonacci sequence modulo 8 has a period of 12: (1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0), the residues are {0, 1, 2, 3, 5, 7} and 4 is not among them. Thus a(7) = -1.
For n = 11: (11 + 1)*(11 + 2*k)/2 = 66 + 12*k . (Fib_i congruent to 6 mod 12) has no solution because the Fibonacci sequence modulo 12 has a period of 24: (1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0), the residues are {0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11} and 6 is not among them. Thus a(11) = -1.

Michel Marcus, Table of n, a(n) for n = 0..3643
Ctibor O. Zizka, Ratio of i/Z(n+1) vs. n for n = 1 to 500. Z(n+1) = A001177(n+1)

Cf. A000045, A000217, A001175, A001177, A161553, A367420.

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
row(n) = {my(L=List([0]), X=Mod([1, 1; 1, 0], n), I=Mod([1, 0; 0, 1], n), M=X); while(M<>I, M*=X; listput(L, lift(M[2, 2]))); Vec(L);} \\ A161553
a(n) = my(p=(n + 1)*(n + 2*k)/2, x=polcoef(p, 1, 'k), y = polcoef(p, 0, 'k), v=row(x)); if (!vecsearch(Set(v),y % x), return(-1)); my(j=0); while ((denominator(jj=((2*fibonacci(j) - n*(n+1))/(2*(n+1)))) != 1) || (numerator(jj)<0), j++); fibonacci(j); \\ Michel Marcus, Jul 19 2025

Corrected and extended by Michel Marcus, Jul 19 2025

A367477 a(n) is the least k such that all possible modular classes a Fibonacci number can take mod n is seen in the Fibonacci numbers Fibonacci(1)..Fibonacci(k).

Original entry on oeis.org

1, 3, 4, 6, 9, 12, 12, 10, 16, 21, 10, 22, 18, 36, 20, 22, 24, 18, 18, 52, 14, 30, 36, 22, 49, 60, 52, 44, 14, 60, 30, 46, 38, 24, 76, 22, 54, 18, 46, 58, 30, 36, 64, 30, 92, 36, 24, 22, 80, 147, 66, 74, 76, 52, 18, 44, 70, 42, 58, 118, 42, 30, 44, 94, 102, 114, 96

Offset: 1

David A. Corneth, Nov 19 2023

nonn

In verifying if k is in A367420 we only need to look from 1 to a(n) to see if there is a Fibonacci number f that has a remainder of k when dividing by 2*k.

			The remainders of Fibonacci numbers mod 4 (starting at Fibonacci(1) = 1) are 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3. The distinct values are {0, 1, 2, 3}. The least k such that the remainders of Fibonacci numbers mod 4 contain all these values is 6 as the first 6 remainders are 1, 1, 2, 3, 1, 0.

Cf. A000045, A001175, A001177, A189768, A367420.

a(n) = {if(n == 1, return(1));
	my(rems = vector(n^2), v = [1,1]);
	rems[1] = 1;
	for(i = 2, n^2,
		rems[i] = v[2];
		v = [v[2], v[1]+v[2]]%n;
		if(v == [1,1],
			break
		)
	);
	s = Set(rems);
	for(i = 1, #rems,
		s = setminus(s, Set(rems[i]));
		if(#s == 0,
			return(i)
		)
	)
}

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A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)(n + 2k)/2, or a(n) = -1 if no such k exists.

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A367477 a(n) is the least k such that all possible modular classes a Fibonacci number can take mod n is seen in the Fibonacci numbers Fibonacci(1)..Fibonacci(k).

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cp's OEIS Frontend

A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)*(n + 2*k)/2, or a(n) = -1 if no such k exists.

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PARI

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A367477 a(n) is the least k such that all possible modular classes a Fibonacci number can take mod n is seen in the Fibonacci numbers Fibonacci(1)..Fibonacci(k).

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A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)(n + 2k)/2, or a(n) = -1 if no such k exists.