cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A237835 a(n) = n*(Pisano period of n) divided by (Pisano period of n^2).

Original entry on oeis.org

1, 1, 1, 1, 1, 6, 1, 1, 1, 2, 1, 12, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 6, 1, 2, 1, 4, 1, 6, 1, 1, 1, 2, 1, 4, 1, 2, 1, 1, 1, 6, 1, 1, 1, 2, 1, 3, 1, 2, 3, 2, 1, 2, 1, 4, 3, 2, 1, 12, 1, 2, 1, 1, 1, 6, 1, 2, 3, 2, 1, 2, 1, 2, 1, 1, 1, 6, 1, 1, 1, 2, 1, 12, 1, 2, 1
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    pp[1] = 1; pp[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[Fibonacci[k+1], n] == 1, Return[k]]];
    a[n_] := n pp[n]/pp[n^2];
    Array[a, 100] (* Jean-François Alcover, Dec 06 2018 *)
  • PARI
    fibmod(n, m)=((Mod([1, 1; 1, 0], m))^n)[1, 2]
    entry_p(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, c+o]; k++); k
    entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, entry_p(f[i, 1]^f[i, 2]), entry_p(f[i, 1])*f[i, 1]^(f[i, 2] - 1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<
    				

Formula

a(n) = n/A237517(n).

A271782 Smallest n-Wall-Sun-Sun prime.

Original entry on oeis.org

13, 241, 2, 3, 191, 5, 2, 3, 2683
Offset: 2

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Author

Felix Fröhlich, Apr 18 2016

Keywords

Comments

A prime p is a k-Wall-Sun-Sun prime iff p^2 divides F_k(pi_k(p)), where F_k(n) is the k-Fibonacci numbers, i.e., a Lucas sequence of first kind with (P,Q) = (k,-1), and pi_k(p) is the Pisano period of k-Fibonacci numbers modulo p (cf. A001175, A175181-A175185).
For prime p > 2 not dividing k^2 + 4, it is a k-Wall-Sun-Sun prime iff p^2 | F_k(p-((k^2+4)/p)), where ((k^2+4)/p) is the Kronecker symbol.
a(1) would be the smallest Wall-Sun-Sun prime whose existence is an open question.
a(12)..a(16) = 2, 3, 3, 29, 2. a(18)..a(33) = 3, 11, 2, 23, 3, 3, 2, 5, 79, 3, 2, 7, 23, 3, 2, 239. a(36)..a(38) = 2, 7, 17. a(40), a(41) = 2, 3. a(43)..a(46) = 5, 2, 3, 41. - R. J. Mathar, Apr 22 2016
a(17) = 1192625911, a(35) = 153794959, a(39) = 30132289567, a(47)..a(50) = 139703, 2, 3, 3. If they exist, a(11), a(34), a(42) are greater than 10^12. - Max Alekseyev, Apr 26 2016
Column k = 1 of table T(n, k) of k-th n-Wall-Sun-Sun prime (that table is not yet in the OEIS as a sequence). - Felix Fröhlich, Apr 25 2016
From Richard N. Smith, Jul 16 2019: (Start)
a(n) = 2 if and only if n is divisible by 4.
a(n) = 3 if and only if n == 5, 9, 13, 14, 18, 22, 23, 27, 31 (mod 36). (End)

Crossrefs

Programs

  • PARI
    A271782(k) = forprime(p=2,10^8, if( (([0,1;1,k]*Mod(1,p^2))^(p-kronecker(k^2+4,p)))[1,2]==0, return(p);); ); \\ Max Alekseyev, Apr 22 2016, corrected by Richard N. Smith, Jul 16 2019 to include p=2 and p divides k^2+4

Formula

a(4n) = 2.

Extensions

Edited by Max Alekseyev, Apr 25 2016

A253247 Pisano period of A006190(n^2) divided by Pisano period of A006190(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 1, 7, 8, 9, 5, 11, 4, 13, 7, 5, 16, 17, 9, 19, 10, 21, 11, 23, 8, 25, 13, 27, 7, 29, 5, 31, 32, 33, 17, 35, 36, 37, 19, 39, 40, 41, 7, 43, 11, 45, 23, 47, 16, 49, 25, 51, 26, 53, 27, 55, 14, 57, 29, 59, 10, 61, 31, 63, 64, 65, 11, 67, 17, 69, 35, 71, 72
Offset: 1

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Author

Eric Chen, Apr 09 2015

Keywords

Comments

For all n, a(n)|n.
Conjecture: a(n) = 1 only for n = 1 and 6. (This conjecture is true if and only if the generalized Wall's conjecture to A006190 is true.)
If there exists any prime p such that A175182(p^2) = A175182(p), then the conjecture fails.
For any prime p, these three statements are equivalent:
(1) A175182(p^2) = A175182(p).
(2) A006190(p-(p|13)) = 3 (mod p^2).
(3) A006497(p) = 1 (mod p^2).
Since A175182(241^2) = A175182(241) = 484, so the prime 241 is a Wall-Sun-Sun prime to A006190 (Lucas (P, Q) = (3, -1)) and no others < 10^8, so the conjecture is true for all primes < 10^8 except 241.
All of Wall's theorems are true for A175182. For example, let P(n) = A175182(n), p and q are primes, then P(pq) = lcm(P(p), P(q)), and for every prime p, P(p)|(p-1) if (p|13) = 1, P(p)|(2p+2) if (p|13) = -1 (P(13) = 52, which if divisible by 13), while (p|13) is the Legendre symbol, and the fixed points of A175182 are 1, 6, and 12*13^k, k>0.

Crossrefs

Programs

  • Maple
    F := proc(k, n) option remember; if n <= 1 then n; else k*procname(k, n-1)+procname(k, n-2) ; end if; end proc:
    Pper := proc(k, m) local cha, zer, n, fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k, n) mod m ; cha := [op(cha), fmodm] ; if fmodm = 0 then zer := [op(zer), n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2], cha) ] = [ op(zer[2]+1..zer[3], cha) ] and [op(1..zer[2], cha)] = [ op(zer[3]+1..zer[4], cha) ] and [op(1..zer[2], cha)] = [ op(zer[4]+1..zer[5], cha) ] then return zer[2] ; elif [op(1..zer[3], cha) ] = [ op(zer[3]+1..zer[5], cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
    k := 3 ; seq( Pper(k, m^2) div Pper(k, m), m=1..300) ;
  • Mathematica
    A006190[n_] := Fibonacci[n, 3];
    A175182[n_] := Module[{k=1}, While[Mod[A006190[k], n] != 0 || Mod[A006190[k+1]-1, n] != 0, k++]; k];
    Table[A175182[n^2] / A175182[n], {n, 72}] (* corrected by Jason Yuen, Jun 28 2025 *)
  • PARI
    fibmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
    entry_p(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, 3*c+o]; k++); k
    entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e8 && f[i, 1] != 241, entry_p(f[i, 1]^f[i, 2]), entry_p(f[i, 1])*f[i, 1]^(f[i, 2] - 1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<
    				

Formula

a(n) = A175182(n^2) / A175182(n).

A237836 Pisano period of n^2.

Original entry on oeis.org

1, 6, 24, 24, 100, 24, 112, 96, 216, 300, 110, 24, 364, 336, 600, 384, 612, 216, 342, 600, 336, 330, 1104, 96, 2500, 1092, 1944, 336, 406, 600, 930, 1536, 1320, 612, 2800, 216, 2812, 342, 2184, 2400, 1640, 336, 3784, 1320, 5400, 1104, 1504, 384, 5488, 7500
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • PARI
    fibmod(n, m)=((Mod([1, 1; 1, 0], m))^n)[1, 2]
    entry_p(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, c+o]; k++); k
    entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, entry_p(f[i, 1]^f[i, 2]), entry_p(f[i, 1])*f[i, 1]^(f[i, 2] - 1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<
    				

Formula

a(n) = A001175(n^2).
Showing 1-4 of 4 results.