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.

A100991 Duplicate of A075821.

Original entry on oeis.org

0, 1, 3, 4, 7, 8, 9, 11, 12, 13, 16, 17, 19, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 36, 37, 39
Offset: 1

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Author

Keywords

A000993 Number of distinct quadratic residues mod 10^n; also number of distinct n-digit endings of base-10 squares.

Original entry on oeis.org

1, 6, 22, 159, 1044, 9121, 78132, 748719, 7161484, 70800861, 699869892, 6978353179, 69580078524, 695292156201, 6947835288052, 69465637212039, 694529215501164, 6944974263529141, 69446563720728612, 694457689921141299, 6944497426351013404
Offset: 0

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Author

N. J. A. Sloane

Keywords

Examples

			Any square ends with one of 0, 1, 4, 5, 6, 9, so a(1) = 6.
A square may end with 22 different two-digit combinations: 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, 96. E.g., no number ending with 14 can be square, etc. See also A075821, A075823.
The finite sequence A122986 has a(3) = 159 terms. - _Reinhard Zumkeller_, Mar 21 2010

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover Publ., 2nd Ed., NY, 1966, Chapter XV, 'On The Square', p. 139.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A036688, A023105, A039300-A039306, A075821, A075823.

Programs

  • Magma
    [1] cat [(83 + 27*(-1)^n + 9*2^(1 + n) + (-1)^n*2^(2 + n) + 9*5^(2 + n) + (-1)^n*5^(2 + n) + 2^(1 + n)*5^(2 + n))/ 72: n in [0..20]]; // Vincenzo Librandi, Mar 29 2012
  • Maple
    -(-6+38*z+241*z^2-594*z^3-1285*z^4+1600*z^5+1500*z^6)/((-1+z)*(5*z-1)*(2*z+1)*(2*z-1)*(5*z+1)*(10*z-1)*(z+1)); #  Bruno Salvy
  • Mathematica
    a[n_] := (83 - 27*(-1)^n + 9*2^(n) - (-1)^n*2^(1 + n) + 9*5^(1 + n) - (-1)^n*5^(1 + n) + 2^(n)*5^(1 + n))/72; Table[ Floor[ a[n]], {n, 0, 20}]
(* Or *) a[0] = 1; a[1] = 6; a[2] = 22; a[3] = 159; a[4] = 1044; a[5] = 9121; a[6] = 78132; a[7] = 748719; a[8] = 7161484; a[n_] := 130 a[n - 2] - 3129 a[n - 4] + 13000 a[n - 6] - 10000 a[n - 8]; Table[ a[n], {n, 0, 20}]
(* Or *) CoefficientList[ Series[(1 - 4*x - 68*x^2 + 59*x^3 + 723*x^4 - 5*x^5 - 1700*x^6 - 500*x^7)/(1 - 10*x - 30*x^2 + 300*x^3 + 129*x^4 - 1290*x^5 - 100*x^6 + 1000*x^7), {x, 0, 20}], x] (* Robert G. Wilson v, Nov 27 2004 *)
LinearRecurrence[{10,30,-300,-129,1290,100,-1000},{1,6,22,159,1044,9121,78132,748719},20] (* Harvey P. Dale, Dec 17 2017 *)
  • Python
    print([(2 + 2**n // 6) * (1 + 5**(n+1) // 12) if n else 1 for n in range(21)]) # Nick Hobson, Mar 10 2024

Formula

a(n) = floor( (83 - (-1)^n*(27 + 2^(n+1) + 5^(n+1)) + 9*2^n + (9 + 2^n)*5^(n+1)) / 72 ).
a(n+8) = 130 a(n+6) - 3129 a(n+4) + 13000 a(n+2) - 10000 a(n) for n >= 1.
G.f.: (1 - 4*x - 68*x^2 + 59*x^3 + 723*x^4 - 5*x^5 - 1700*x^6 - 500*x^7)/(1 - 10*x - 30*x^2 + 300*x^3 + 129*x^4 - 1290*x^5 - 100*x^6 + 1000*x^7).

A046638 Number of cubic residues mod 10^n, or number of distinct n-digit endings of cubes.

Original entry on oeis.org

1, 10, 63, 505, 5050, 47899, 466237, 4662370, 46308087, 461504593, 4615045930, 46111077091, 460913873941, 4609138739410, 46086465166623, 460840040641225, 4608400406412250, 46083388790070379, 460830811531341997, 4608308115313419970, 46083004243912737927
Offset: 0

Views

Author

David W. Wilson

Keywords

Examples

			a(1)=10 because a cube may end with any digit (10 possible combinations); a(2)=63 because a cube may end with 63 2-digit combinations (including leading zeros).
A cube may end with 63 different 2-digit combinations: 00, 01, 03, 04, 07, 08, 09, 11, 12, 13, 16, 17, 19, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 36, 37, 39, 41, 43, 44, 47, 48, 49, 51, 52, 53, 56, 57, 59, 61, 63, 64, 67, 68, 69, 71, 72, 73, 75, 76, 77, 79, 81, 83, 84, 87, 88, 89, 91, 92, 93, 96, 97, 99. Numbers ending with 14 say cannot be cubes. See also A075821, A075823. - _Zak Seidov_, Oct 18 2002

Links

Crossrefs

Cf. A075821, A075823.

Programs

  • PARI
    a(n)=(5^(n+2)+30)\31*((4<Charles R Greathouse IV, Jan 03 2013

Formula

a(n) = A046530(10^n) = A046630(n)*A046633(n). - R. J. Mathar, Feb 28 2011
a(n) ~ 100/217 * 10^n, so large terms start 460829493.... - Charles R Greathouse IV, Jan 03 2013
G.f.: -(10000*x^9+9000*x^8-5130*x^6-2357*x^5+259*x^3+37*x^2-1) / ((x-1)*(2*x-1)*(5*x-1)*(10*x-1)*(x^2+x+1)*(25*x^2+5*x+1)*(4*x^2+2*x+1)). - Alois P. Heinz, Jan 03 2013

Extensions

Edited by N. J. A. Sloane, Oct 19 2008

A100990 a(n) = n^21 mod 100.

Original entry on oeis.org

0, 1, 52, 3, 4, 25, 56, 7, 8, 9, 0, 11, 12, 13, 64, 75, 16, 17, 68, 19, 0, 21, 72, 23, 24, 25, 76, 27, 28, 29, 0, 31, 32, 33, 84, 75, 36, 37, 88, 39, 0, 41, 92, 43, 44, 25, 96, 47, 48, 49, 0, 51, 52, 53, 4, 75, 56, 57, 8, 59, 0, 61, 12, 63, 64, 25, 16, 67, 68, 69, 0, 71, 72, 73, 24
Offset: 0

Views

Author

Henry Bottomley, Nov 25 2004

Keywords

Comments

Also n^(20k+1) mod 100 for any positive integer k.
There are 63 numbers (A075821) where the final two digits of n^21, n^41, n^61, etc. are equal to n.
Period 100.

Examples

			a(11) = 11 since 11^21 = 7400249944258160101211 and the final two digits are 11.

Links

Crossrefs

Cf. A010809, A051126, A075821.

Programs

Formula

a(n) = A051126(A010809(n), 100) = a(n-100).
Showing 1-4 of 4 results.

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

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