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 1-5 of 5 results.

A119690 n! mod n*(n+1)/2.

Original entry on oeis.org

0, 2, 0, 4, 0, 6, 0, 0, 0, 10, 0, 12, 0, 0, 0, 16, 0, 18, 0, 0, 0, 22, 0, 0, 0, 0, 0, 28, 0, 30, 0, 0, 0, 0, 0, 36, 0, 0, 0, 40, 0, 42, 0, 0, 0, 46, 0, 0, 0, 0, 0, 52, 0, 0, 0, 0, 0, 58, 0, 60, 0, 0, 0, 0, 0, 66, 0, 0, 0, 70, 0, 72, 0, 0, 0, 0, 0, 78, 0, 0, 0, 82, 0, 0, 0, 0, 0, 88, 0, 0, 0, 0, 0, 0
Offset: 1

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Author

Paolo P. Lava and Giorgio Balzarotti, Jun 09 2006

Keywords

Comments

All terms are even.
It appears that f(n)=(n!)^(2k+1) modulo n(n+1)/2 is n if n is one less than an odd prime, else f(n) is 0, for any integer k. See A175567 for related results involving an even power of n!. - John W. Layman, Jul 12 2010

Crossrefs

Cf. A006093, A010051, A175567, A175553.

Programs

  • Magma
    [ Factorial(n) mod (n*(n+1) div 2): n in [1..100] ];
  • Maple
    P:=proc(n) local i, j, k; j:=1; k:=0; for i from 1 by 1 to n do j:=j*i; k:=k+i; print(j mod k); od; end: P(100);

Formula

a(n) = n if n+1 is an odd prime, a(n) = 0 otherwise.

A133653 A007318^(-1) * A003261.

Original entry on oeis.org

1, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154
Offset: 1

Views

Author

Gary W. Adamson, Sep 19 2007

Keywords

Comments

It appears this sequence gives the positive integers m such that the sum of the first m Fibonacci numbers divides their product. For example, if n=2 and m=a(2)=6, we have the sum 1+1+2+3+5+8=20 which clearly divides the corresponding product 480. See A175553 for the analogous sequence when using the triangular numbers. Sum_{k=1..n} Fibonacci(k) divides Product_{k=1..n} Fibonacci(k). - John W. Layman, Jul 10 2010

Examples

			a(4) = 14 = (1, 3, 3, 1) dot (1, 5, -1, 1) = (1, 15, -3, 1).

Crossrefs

Cf. A003261, A007318, A175553.
Essentially the same as A130824, A113127, A111284, A073760, A016825.

Formula

Inverse binomial transform of A003261: (1, 7, 23, 63, 159, 383, ...).
Binomial transform of [1, 5, -1, 1, -1, 1, ...].
"1" followed by 2 * [3, 5, 7, 9, 11, ...].
O.g.f.: x*(1+4x-x^2)/(1-x)^2. a(n) = 4n-2, n > 1. - R. J. Mathar, Jun 08 2008
1/(1+1/(6+1/(10+1/(14+1/(...(continued fraction)))))) = (e-1)/2 with e = 2.718281...- Philippe Deléham, Mar 09 2013

Extensions

More terms from R. J. Mathar, Jun 08 2008

A175567 (n!)^2 modulo n(n+1)/2.

Original entry on oeis.org

0, 1, 0, 6, 0, 15, 0, 0, 0, 45, 0, 66, 0, 0, 0, 120, 0, 153, 0, 0, 0, 231, 0, 0, 0, 0, 0, 378, 0, 435, 0, 0, 0, 0, 0, 630, 0, 0, 0, 780, 0, 861, 0, 0, 0, 1035, 0, 0, 0, 0, 0, 1326, 0, 0, 0, 0, 0, 1653, 0, 1770, 0, 0, 0, 0, 0, 2145, 0, 0, 0, 2415, 0, 2556, 0, 0, 0, 0, 0, 3003, 0, 0, 0, 3321
Offset: 1

Views

Author

John W. Layman, Jul 12 2010

Keywords

Comments

It appears that if n is one less than an odd prime then (n!)^2 modulo n(n+1)/2 is n(n-1)/2 else 0. This result appears to hold for any even power of n!. See A119690 for similar results related to odd powers of n!.

Links

Crossrefs

Cf. A119690, A169690, A175553.

Programs

  • Mathematica
    Table[Mod[(n!)^2, (n^2 + n)/2], {n, 100}] (* Vincenzo Librandi, Jul 10 2014 *)
Table[PowerMod[n!,2,(n(n+1))/2],{n,100}] (* Harvey P. Dale, Aug 27 2016 *)
  • PARI
    a(n) = (n!)^2 % (n*(n+1)/2); \\ Michel Marcus, Jul 09 2014

A159951 Fibonacci integral quotients associated with the dividends in A159950 and the divisors in A003481.

Original entry on oeis.org

12, 856800, 139890541190400, 50664770469826998541056000, 40527253814267058837705250384270510080000, 71554565901386985191123530075861409411081105273676595200000
Offset: 1

Views

Author

Enoch Haga, Apr 27 2009

Keywords

Comments

The first example of an integral quotient in the Fibonacci sequence is 12 because 240/20=12. 240 is the product of terms through 8, and 20 the sum. Thereafter, with every other additional pair of terms in the Fibonacci sequence, another integral quotient occurs.
Let m be an even positive integer. Then the sequence defined by b_m(n) = Product_{k = 1..2*n+1} F(m*k) / Sum_{k = 1..2*n+1} F(m*k) appears to be integral. - Peter Bala, Nov 12 2021

Examples

			The first two integral quotients occur in the Fibonacci sequence as illustrated by the following: (1*1*2*3*5*8)/(1+1+2+3+5+8) = 240/20 = 12, integral; (1*1*2*3*5*8*13*21*34*55)/(1+1+2+3+5+8+13+21+34+55) = 122522400/143 = 856800, integral.

Crossrefs

Cf. A000045, A001519, A001906, A003481, A033890, A159950, A175553.

Programs

  • Maple
    with(combinat):
seq(mul(fibonacci(k), k = 1..4*n+2)/(fibonacci(4*n+4) - 1), n = 1..10); # Peter Bala, Nov 04 2021
  • UBASIC
    10 'Fibo 20 'R=SUM:S=PRODUCT 30 'T integral every other pair 40 A=1:S=1:print A;:S=S*1 50 B=1:print B;:S=S*B 60 C=A+B:print C;:R=R+C:S=S*C 70 D=B+C:print D;:R=R+D:R=R+2:print R:S=S*D:print S 80 T=S/R:if T=int(S/R) then print T:stop 90 A=C:B=D:R=R-2:goto 60

Formula

a(n) = (Product_{k = 1..4*n+2} Fibonacci(k))/(Sum_{k = 1..4*n+2} Fibonacci(k)) = (Product_{k = 1..4*n+2} Fibonacci(k))/(Fibonacci(4*n+4) - 1) = Fibonacci(2*n+1)/Fibonacci(2*n+3) * Product_{k = 1..4*n+1} Fibonacci(k), which shows a(n) is integral. Cf. A175553. - Peter Bala, Nov 11 2021

A349272 a(n) = Product_{k = 1..2*n+1} Fibonacci(2*k) / Sum_{k = 1..2*n+1} Fibonacci(2*k).

Original entry on oeis.org

1, 2, 315, 2471040, 918185538816, 16047302734562299200, 13178031727820369629763174400, 508406658175888466343652105865846784000, 921456090985190879093613420564815806955580862464000, 78458394721620642094151397745899367347021362840662985785265356800
Offset: 0

Views

Author

Peter Bala, Nov 12 2021

Keywords

Comments

Let m be an even positive integer. We conjecture that the sequence defined by b_m(n) = Product_{k = 1..2*n+1} Fibonacci(m*k) / Sum_{k = 1..2*n+1} Fibonacci(m*k) is integral. The formula given below proves the conjecture in the present case m = 2. The cases m = 4 and m = 6 of the conjecture can be proved in a similar manner.
More generally, if F(n,x) denotes the n-th Fibonacci polynomial we conjecture that, for each n, the rational function Product_{k = 1..2*n+1} F(m*k,x) / Sum_{k = 1..2*n+1} F(m*k,x) is an integral polynomial.

Links

Crossrefs

Cf. A000045, A159951, A175553.
Cf. A003266, A194157.

Programs

  • Maple
    with(combinat):
seq(mul(fibonacci(2*k), k = 1..2*n+1)/add(fibonacci(2*k), k = 1..2*n+1), n = 0..10);
  • Mathematica
    Table[Product[ Fibonacci[2k],{k,2n+1}]/Sum[Fibonacci[2k],{k,2n+1}],{n,0,9}] (* Stefano Spezia, Nov 13 2021 *)
  • PARI
    a(n) = prod(k = 1, 2*n+1, fibonacci(2*k)) / sum(k = 1, 2*n+1, fibonacci(2*k)); \\ Michel Marcus, Nov 12 2021

Formula

a(n) = F(2*n+1)/F(2*n+2) * Product_{k = 1..2*n} Fibonacci(2*k), shows a(n) to be integral. Cf. A159951.
a(n) ~ A194159 * phi^(4*n^2 + 2*n - 1) / 5^n, where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 31 2023
Showing 1-5 of 5 results.

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