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.

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A075414 Squares of A002279: a(n) = (5*(10^n - 1)/9)^2.

Original entry on oeis.org

0, 25, 3025, 308025, 30858025, 3086358025, 308641358025, 30864191358025, 3086419691358025, 308641974691358025, 30864197524691358025, 3086419753024691358025, 308641975308024691358025, 30864197530858024691358025, 3086419753086358024691358025, 308641975308641358024691358025
Offset: 0

Views

Author

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

Keywords

Comments

A transformation of the Wonderful Demlo numbers (A002477).

Examples

			a(2) = 55^2 = 3025.
		

Crossrefs

Programs

  • PARI
    concat(0, Vec(25*x*(1 + 10*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20))) \\ Colin Barker, Jul 17 2019

Formula

a(n) = A002279(n)^2 = (5*A002275(n))^2 = 25*A002275(n)^2.
From Colin Barker, Jul 17 2019: (Start)
G.f.: 25*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2.
a(n) = 25*(10^n-1)^2/81. (End)
From Elmo R. Oliveira, Jul 29 2025: (Start)
E.g.f.: 25*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 25*A002477(n). (End)

A075416 Squares of A002281.

Original entry on oeis.org

0, 49, 5929, 603729, 60481729, 6049261729, 604937061729, 60493815061729, 6049382595061729, 604938270395061729, 60493827148395061729, 6049382715928395061729, 604938271603728395061729, 60493827160481728395061729, 6049382716049261728395061729, 604938271604937061728395061729
Offset: 0

Views

Author

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

Keywords

Comments

A transformation of the Wonderful Demlo numbers (A002477).

Examples

			a(2) = 77^2 = 5929.
		

Crossrefs

Programs

  • PARI
    concat(0, Vec(49*x*(1 + 10*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20))) \\ Colin Barker, Jul 17 2019

Formula

a(n) = A002281(n)^2 = (7*A002275(n))^2 = 49*A002275(n)^2.
From Colin Barker, Jul 17 2019: (Start)
G.f.: 49*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2.
a(n) = 49*(10^n-1)^2/81. (End)
From Elmo R. Oliveira, Jul 29 2025: (Start)
E.g.f.: 49*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 49*A002477(n). (End)

A075417 Squares of A002282: a(n) = (8*(10^n - 1)/9)^2.

Original entry on oeis.org

0, 64, 7744, 788544, 78996544, 7901076544, 790121876544, 79012329876544, 7901234409876544, 790123455209876544, 79012345663209876544, 7901234567743209876544, 790123456788543209876544, 79012345678996543209876544, 7901234567901076543209876544, 790123456790121876543209876544
Offset: 0

Views

Author

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

Keywords

Comments

A transformation of the Wonderful Demlo numbers (A002477).

Examples

			a(2) = 88^2 = 7744.
		

Crossrefs

Programs

  • PARI
    concat(0, Vec(64*x*(1 + 10*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20))) \\ Colin Barker, Jul 17 2019

Formula

a(n) = A002282(n)^2 = (8*A002275(n))^2 = 64*A002275(n)^2.
From Colin Barker, Jul 17 2019: (Start)
G.f.: 64*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2.
a(n) = 64*(10^n-1)^2/81. (End)
From Elmo R. Oliveira, Jul 30 2025: (Start)
E.g.f.: 64*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 64*A002477(n). (End)

A075024 a(n) is the largest prime divisor of the number A173426(n) = concatenate(1,2,...,n-1,n,n-1,...,2,1); a(1) = 1.

Original entry on oeis.org

1, 11, 37, 101, 271, 37, 4649, 137, 333667, 12345678910987654321, 17636684157301569664903, 2799473675762179389994681, 2354041513534224607850261, 2068140300159522133, 498056174529497, 112240064764214229701, 4188353169004802474320231191377
Offset: 1

Views

Author

Amarnath Murthy, Sep 01 2002

Keywords

Comments

Also for 1 < n < 10, a(n) is the common prime divisor for all A010785(m) which consist of n digits. - Alexander R. Povolotsky, Jun 05 2014, corrected by M. F. Hasler, Jul 30 2015
According to the definition (and given terms), this is the greatest prime factor (A006530) of A173426 and not of A002477, as an earlier formula asserted and which may have been an assumption of the preceding comment. - M. F. Hasler, Jul 29 2015

Examples

			a(5) = 271 as 123454321 = 41*41*271*271.
a(25) = 12471243489559387823527232424981012432152516319410549 is the larger factor of the semiprime A173426(24) = A075023(25) * a(n).
		

Crossrefs

Programs

  • Mathematica
    Table[FactorInteger[FromDigits[Join[Flatten[IntegerDigits/@Range[ n]], Flatten[ IntegerDigits/@Range[n-1,1,-1]]]]][[-1,1]],{n,20}] (* Harvey P. Dale, May 20 2016 *)
  • PARI
    a(n) = {if (n == 1, return (1)); s = ""; for (i=1, n, s = concat(s, Str(i));); forstep (i=n-1, 1, -1, s = concat(s, Str(i));); f = factor(eval(s)); f[#f~, 1];} \\ Michel Marcus, Jun 05 2014
    
  • PARI
    A075024(n)=A006530(A173426(n)) \\ A006530 should provide efficient code and also covers the case n=1. - M. F. Hasler, Jul 29 2015

Formula

a(n) = A006530(A173426(n)). - Michel Marcus, Jun 05 2014, corrected by M. F. Hasler, Jul 29 2015

Extensions

More terms from Sascha Kurz, Jan 03 2003
a(16)-a(17) from Michel Marcus, Jun 05 2014
More terms from M. F. Hasler, Jul 29 2015

A057139 Odd number of digits palindrome based on sequential digits.

Original entry on oeis.org

1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321, 1234567890987654321, 123456789010987654321, 12345678901210987654321, 1234567890123210987654321, 123456789012343210987654321, 12345678901234543210987654321
Offset: 1

Views

Author

Henry Bottomley, Aug 12 2000

Keywords

Crossrefs

Alternative progression for n >= 10 compared with A002477.

Programs

  • Mathematica
    Array[FromDigits@ Join[#, Reverse@ Most@ #] &@ Mod[Range[#], 10] &, 15] (* Michael De Vlieger, Jan 28 2020 *)
  • PARI
    a(n)={fromdigits(vector(2*n-1, i, if(i<=n, i, 2*n-i)%10))} \\ Andrew Howroyd, Jan 27 2020

Formula

a(n) = 10^n*A057137(n-1) + A057138(n) = 10^(n-1)*A057137(n) + A057138(n-1).

Extensions

Terms a(13) and beyond from Andrew Howroyd, Jan 27 2020

A048411 Squares whose consecutive digits differ by 1.

Original entry on oeis.org

0, 1, 4, 9, 121, 676, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321
Offset: 1

Views

Author

Patrick De Geest, Apr 15 1999

Keywords

Comments

a(14), if it exists, is > 10^34. - Lars Blomberg, Nov 25 2016
Is it true that all terms are palindromes? - Chai Wah Wu, Apr 06 2018
a(14), if it exists, is > 10^52. - Michael S. Branicky, Apr 22 2025

Crossrefs

Cf. A010052; intersection of A033075 and A000290.

Programs

  • Haskell
    a048411 n = a048411_list !! (n-1)
    a048411_list = filter ((== 1) . a010052) a033075_list
    -- Reinhard Zumkeller, Feb 21 2012
    
  • Mathematica
    Select[Range[0, 10^7]^2, Or[# == 0, IntegerLength@ # == 1, Union@ Abs@ Differences@ IntegerDigits@ # == {1}] &] (* Michael De Vlieger, Nov 25 2016 *)
  • Python
    from sympy.ntheory.primetest import is_square
    def gen(d, s=None):
        if d == 0: yield tuple(); return
        if s == None:
            yield from [(i, ) + g for i in range(1, 10) for g in gen(d-1, s=i)]
        else:
            if s > 0: yield from [(s-1, ) + g for g in gen(d-1, s=s-1)]
            if s < 9: yield from [(s+1, ) + g for g in gen(d-1, s=s+1)]
    def afind(maxdigits):
        print(0, end=", ")
        for d in range(1, maxdigits+1):
            for g in gen(d, s=None):
                t = int("".join(map(str, g)))
                if is_square(t): print(t, end=", ")
    afind(17) # Michael S. Branicky, Sep 26 2021

Formula

a(n) = A048412(n)^2.

A249605 Dissectible numbers in the sense of Gunjikar and Kaprekar.

Original entry on oeis.org

9, 18, 27, 36, 45, 54, 63, 72, 81, 108, 117, 126, 135, 144, 153, 162, 207, 216, 225, 234, 243, 306, 315, 324, 405
Offset: 1

Views

Author

N. J. A. Sloane, Nov 04 2014

Keywords

Comments

The numbers really have exactly three digits, allowing leading zeros. The full list is: 009, 018, 027, 036, 045, 054, 063, 072, 081, 108, 117, 126, 135, 144, 153, 162, 207, 216, 225, 234, 243, 306, 315, 324, 405.
A dissectible number is a three-digit number abc with the property that when multiplied by any Wonderful Demlo number (A002477) the product has the form axx..xbyy..yc, with the same number of x digits as y digits.

Examples

			9*12321 = 009*12321 = 0110889 (a=b=0, c=9, x=1, y=8).
162*121 = 19602 (here x=9, y=0).
162*12321 = 1996002 (again x=9, y=0).
		

Crossrefs

Cf. A002477.

A048412 a(n)^2 is a square whose consecutive digits differ by 1.

Original entry on oeis.org

0, 1, 2, 3, 11, 26, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111
Offset: 1

Views

Author

Patrick De Geest, Apr 15 1999

Keywords

Comments

a(14), if it exists, is > 10^17. - Lars Blomberg, Nov 25 2016
a(14), if it exists, is > 10^26. - Michael S. Branicky, Apr 22 2025

Crossrefs

Formula

a(n) = sqrt(A048411(n)).

Extensions

Offset changed to 1 by Jinyuan Wang, Sep 04 2021

A093140 Expansion of (1-6*x)/((1-x)*(1-10*x)).

Original entry on oeis.org

1, 5, 45, 445, 4445, 44445, 444445, 4444445, 44444445, 444444445, 4444444445, 44444444445, 444444444445, 4444444444445, 44444444444445, 444444444444445, 4444444444444445, 44444444444444445, 444444444444444445, 4444444444444444445, 44444444444444444445, 444444444444444444445
Offset: 0

Views

Author

Paul Barry, Mar 24 2004

Keywords

Comments

Second binomial transform of 4*A001045(3n)/3+(-1)^n. Partial sums of A093141. A convex combination of 10^n and 1. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is 1, 1+k, 1+11k, 1+111k, ... This is the case for k=4.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-6x)/((1-x)(1-10x)),{x,0,30}],x] (* or *) LinearRecurrence[{11,-10},{1,5},30] (* or *) Join[{1},Table[FromDigits[PadLeft[{5},n,4]],{n,30}]] (* Harvey P. Dale, Dec 17 2022 *)

Formula

G.f.: (1-6*x)/((1-x)*(1-10*x)).
a(n) = 4*10^n/9 + 5/9.
a(n+1) = (A102807(n+1)-A002477(n))/((Sum_{i=1..n} 2*10^i) + 3). [Roger L. Bagula, May 22 2010]
a(n) = 10*a(n-1)-5 with a(0)=1. - Vincenzo Librandi, Aug 02 2010
a(n) = 11*a(n-1)-10*a(n-2). - Wesley Ivan Hurt, May 20 2021
E.g.f.: exp(x)*(4*exp(9*x) + 5)/9. - Elmo R. Oliveira, Aug 17 2024

Extensions

a(19)-a(22) from Elmo R. Oliveira, Aug 17 2024

A180027 Partial sums of A100706.

Original entry on oeis.org

1, 112, 11223, 1122334, 112233445, 11223344556, 1122334455667, 112233445566778, 11223344556677889, 1122334455667789000, 112233445566778900111, 11223344556677890011222, 1122334455667789001122333, 112233445566778900112233444, 11223344556677890011223344555, 1122334455667789001122334455666
Offset: 0

Views

Author

Mark Dols, Aug 07 2010

Keywords

Comments

Up to n=8 the digits of a(n) sum up to n^2.
Similar to this, A014824 (1,12,123,1234,...) is a representation of the triangular numbers; (1,1112,1112223,1112223334,...) of the pentagonal numbers;(1,11112,111122223,...) of the hexagonal numbers, and so on. A nice thing about this sequence(s) is that the (represented) value of the integer matches the partial sums of the number of digits in the sequence.
f(n) = 100*f(n-1) + A100706(n) gives a mirrored version of this sequence, and f(n) = 10*f(n-1) + A100706(n) the symmetrical version (A002477).

Crossrefs

Programs

Formula

a(n) = Sum_{k=0..n} A100706(k). - Michel Marcus, Mar 12 2023

Extensions

More terms and edited by Michel Marcus, Mar 12 2023
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