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 1-3 of 3 results.

A308215 a(n) is the multiplicative inverse of A001844(n+1) modulo A001844(n); where A001844 is the sequence of centered square numbers.

Original entry on oeis.org

0, 2, 12, 11, 39, 28, 82, 53, 141, 86, 216, 127, 307, 176, 414, 233, 537, 298, 676, 371, 831, 452, 1002, 541, 1189, 638, 1392, 743, 1611, 856, 1846, 977, 2097, 1106, 2364, 1243, 2647, 1388, 2946, 1541, 3261, 1702, 3592, 1871, 3939, 2048
Offset: 0

Views

Author

Daniel Hoyt, May 15 2019

Keywords

Comments

The sequence explores the relationship between the terms of A001844, the sums of consecutive squares. The sequence is an interleaving of A054552 (a number spiral arm) and (A001844-n). The gap between the lower values of A308215 and the upper values of A308217 increase by 3n; each successive gap increasing by 6.

Links

Crossrefs

Cf. A001844, A033951, A054552, A308217.

Programs

  • PARI
    f(n) = 2*n*(n+1)+1; \\ A001844
a(n) = lift(1/Mod(f(n+1), f(n))); \\ Michel Marcus, May 16 2019
  • Python
    import gmpy2
sos = [] # sum of squares
a=0
b=1
for i in range(50):
    c = a**2 + b**2
    sos.append(c)
    a +=1
    b +=1
ls = []
for i in range(len(sos)-1):
    c = gmpy2.invert(sos[i+1],sos[i])
    ls.append(int(c))
print(ls)

Formula

a(n) satisfies a(n)*(2*n*(n+1)+1) == 1 (mod 2*n*(n-1)+1).
Conjectures from Colin Barker, May 16 2019: (Start)
G.f.: x*(2 + 12*x + 5*x^2 + 3*x^3 + x^4 + x^5) / ((1 - x)^3*(1 + x)^3).
a(n) = (3 + (-1)^n + 2*(2+(-1)^n)*n + 2*(3+(-1)^n)*n^2) / 4 for n>0.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>6.
(End)

A334121 a(n) is the multiplicative inverse of A008514(n) modulo A008514(n+1).

Original entry on oeis.org

1, 40, 271, 183, 1184, 102, 2861, 9102, 4446, 16837, 3483, 25332, 62253, 30739, 88410, 26394, 111803, 2177, 123512, 298353, 110585, 346624, 55398, 366215, 782616, 335806, 866645, 228189, 893798, 9628, 832307, 1873716, 643300, 1905211, 280161, 1793338, 3647187
Offset: 0

Views

Author

Daniel Hoyt, Apr 14 2020

Keywords

Crossrefs

Cf. A008514, A308215, A308217, A334137.

Programs

  • PARI
    f(n) = n^4 + (n+1)^4; \\ A008514
a(n) = lift(1/Mod(f(n), f(n+1)));
  • Python
    import gmpy2
soc = [] # sum of 4d-centered cubes
a=0
b=1
for i in range(100):
    c = a**4 + b**4
    soc.append(c)
    a += 1
    b += 1
A334121 = []
for i in range(len(soc)-1):
    c = gmpy2.invert(soc[i], soc[i+1])
    A334121.append(int(c))
print(', '.join([str(x) for x in A334121]))

A334137 a(n) is the multiplicative inverse of A008514(n+1) modulo A008514(n).

Original entry on oeis.org

0, 10, 19, 267, 338, 1868, 2069, 948, 7796, 5245, 22215, 17198, 3477, 43855, 21272, 95592, 60647, 186753, 135194, 45969, 263049, 139666, 467532, 301563, 55386, 559636, 241005, 948261, 543212, 1508854, 1001903, 318414, 1664590, 828391, 2587041, 1575400, 280143
Offset: 0

Views

Author

Daniel Hoyt, Apr 15 2020

Keywords

Examples

			For a(0), compute the inverse of 1^4 + 2^4 mod 0^4 + 1^4 which is 0 mod 1, since everything mod 1 is 0.
For a(1), compute the inverse of 2^4 + 3^4 mod 1^4 + 2^4. The inverse of 97 mod 17 (or 12 mod 17) is 10 mod 17 since 10*12 = 120 has remainder 1 mod 17.

Links

Crossrefs

Cf. A308215, A308217, A008514, A334121.

Programs

  • PARI
    f(n) = n^4 + (n+1)^4;
a(n) = lift(1/Mod(f(n+1), f(n)));
  • Python
    import gmpy2
soc = [] # sum of 4d-centered cubes
a=0
b=1
for i in range(100):
    c = a**4 + b**4
    soc.append(c)
    a += 1
    b += 1
A334137 = []
for i in range(len(soc)-1):
    c = gmpy2.invert(soc[i+1], soc[i])
    A334137.append(int(c))
print(', '.join([str(x) for x in A334137]))
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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