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 1-10 of 35 results. Next

A080151 Let m = Wonderful Demlo number A002477(n); a(n) = sum of digits of m.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 81, 82, 85, 90, 97, 106, 117, 130, 145, 162, 163, 166, 171, 178, 187, 198, 211, 226, 243, 244, 247, 252, 259, 268, 279, 292, 307, 324, 325, 328, 333, 340, 349, 360, 373, 388, 405, 406, 409, 414, 421, 430, 441, 454, 469, 486, 487
Offset: 1

Views

Author

Eric W. Weisstein, Jan 31 2003

Keywords

Comments

Record values in A003132. - Reinhard Zumkeller, Jul 10 2011

Crossrefs

Programs

  • Haskell
    a n=(div n 9)*81+(mod n 9)^2
              A080151=map a [1..] \\ Chernin Nadav, Mar 06 2014
    
  • Maple
    f := n -> 9*n - 81*frac(1/9*n) + 81*frac(1/9*n)^2:
    map(f, [$1..100]); # Robert Israel, Aug 05 2019
  • Mathematica
    (* by direct counting *)
    Repunit[n_] := (-1 + 10^n)/9; A080151[n_]:=Plus @@ IntegerDigits[Repunit[n]^2];
    (* by the formula *)
    A080151[n_] := (9^2)*(n/9 - FractionalPart[n/9] + FractionalPart[n/9]^2)
    (* or alternatively *)
    A080151[n_] := 81*(Floor[n/9]+ FractionalPart[n/9]^2) (* Enrique Pérez Herrero, Nov 22 2009 *)
  • PARI
    vector(100, n, (n\9)*81+(n%9)^2) \\ Colin Barker, Mar 05 2014

Formula

a(n) = A007953(A002477(n)).
a(n) = sqrt( A080150(n) ).
a(n) = (9^2)*(n/9 - {n/9} + {n/9}^2) = 81*(floor(n/9) + {n/9}^2), where the symbol {n} means fractional part of n. - Enrique Pérez Herrero, Nov 22 2009
a(n) = A003132(A051885(n)). - Reinhard Zumkeller, Jul 10 2011
a(9*n + k) = 81*n + k^2, with k in range 0 to 9. - Enrique Pérez Herrero, Nov 05 2022
Empirical g.f.: x*(17*x^8 + 15*x^7 + 13*x^6 + 11*x^5 + 9*x^4 + 7*x^3 + 5*x^2 + 3*x + 1) / ((x-1)^2*(x^2+x+1)*(x^6+x^3+1)). - Colin Barker, Mar 05 2014
Empirical g.f. confirmed. - Robert Israel, Aug 05 2019

A080160 Squares that are digit sums of Wonderful Demlo numbers A002477.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 81, 324, 441, 576, 729, 900, 1089, 1296, 2025, 2304, 2601, 2916, 3249, 3600, 3969, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 9801, 10404, 11025, 11664
Offset: 1

Views

Author

N. J. A. Sloane, Jun 19 2005

Keywords

Comments

These are the squares in A080151.
x^2 is a term iff x mod 81 is in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 18, 21, 24, 27, 30, 33, 36, 45, 48, 51, 54, 57, 60, 63, 72, 73, 74, 75, 76, 77, 78, 79, 80}. - Robert Israel, Jul 30 2025

Crossrefs

Programs

  • Maple
    b:=n->sum(convert(((10^(n+1)-1)/9)^2,base,10)[j],j=1..2*n+1): a:=proc(n) if type(sqrt(b(n)),integer)=true then b(n) else fi end: seq(a(n),n=0..2000); # Emeric Deutsch, Jun 19 2005
  • Mathematica
    A080151[n_] := (9^2)*(n/9 - FractionalPart[n/9] + FractionalPart[n/9]^2)
    A080151[Select[Range[10000], IntegerQ[Sqrt[A080151[#]]] &]]
    (* Enrique Pérez Herrero, Nov 05 2022 *)

Extensions

More terms from Emeric Deutsch, Jun 19 2005

A080161 Indices of Wonderful Demlo numbers A002477 whose digit sums are squares.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 36, 51, 66, 81, 102, 123, 144, 225, 258, 291, 324, 363, 402, 441, 576, 593, 610, 627, 644, 661, 678, 695, 712, 729, 748, 767, 786, 805, 824, 843, 862, 881, 900, 1089, 1158, 1227, 1296, 1371, 1446, 1521, 1764, 1851, 1938, 2025
Offset: 1

Views

Author

Eric W. Weisstein, Jan 31 2003

Keywords

Comments

The numbers 9*n^2 (A016766), with n > 0, are in this sequence. - Enrique Pérez Herrero, Sep 26 2020

Crossrefs

Programs

  • Mathematica
    A080151[n_] := (9^2)*(n/9 - FractionalPart[n/9] + FractionalPart[n/9]^2)
    Select[Range[10000], IntegerQ[Sqrt[A080151[#]]] &]
    (* Enrique Pérez Herrero, Sep 26 2020 *)
  • PARI
    for(k=1,10^5,issquare((k\9)*81+(k%9)^2)&&print1(k,", ")) \\ Jeppe Stig Nielsen, May 27 2023

A080162 Wonderful Demlo numbers A002477 whose digit sums are squares.

Original entry on oeis.org

1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321, 12345679012345679012345679012345678987654320987654320987654320987654321
Offset: 1

Views

Author

Eric W. Weisstein, Jan 31 2003

Keywords

Comments

The next term (a(11)) has 101 digits. - Harvey P. Dale, Jun 16 2025

Crossrefs

Programs

  • Mathematica
    Select[LinearRecurrence[{111,-1110,1000},{1,121,12321},40],IntegerQ[Sqrt[Total[IntegerDigits[#]]]]&] (* Harvey P. Dale, Jun 16 2025 *)
  • PARI
    for(k=1,100,my(d=((10^k-1)/9)^2); issquare(sumdigits(d)) && print1(d,", ")) \\ Jeppe Stig Nielsen, May 27 2023

A063750 Number of divisors (A000005) of the Wonderful Demlo numbers A002477.

Original entry on oeis.org

1, 3, 9, 9, 9, 243, 9, 81, 45, 81, 9, 2187, 27, 81, 729, 729, 9, 10935, 3, 2187, 2187, 1215, 3, 59049, 243, 729, 567, 6561, 243, 1594323, 27, 177147, 729, 729, 2187, 295245, 27, 27, 729, 177147, 81, 7971615, 81, 98415, 32805, 729
Offset: 1

Views

Author

Jason Earls, Aug 11 2001

Keywords

Crossrefs

Programs

Formula

a(n) = A000005(A002477(n)) = d(((10^n-1)/9)^2). - M. F. Hasler, Nov 18 2017
a(n) = A048691(A000042(n)). - Amiram Eldar, Nov 19 2024

Extensions

Edited and offset corrected by M. F. Hasler, Nov 18 2017

A063751 a(n) = phi(A002477(n)).

Original entry on oeis.org

1, 110, 7992, 1111000, 119998800, 5759994240, 1229137654864, 108799998912000, 8007983991992016, 1090799999890920000, 123450846932098824864, 5702399999994297600000, 1195941578346547072508832, 111739686239998882603137600, 7532766719999992467233280000
Offset: 1

Views

Author

Jason Earls, Aug 11 2001

Keywords

Crossrefs

Programs

  • Mathematica
    EulerPhi[((10^Range[20] - 1)/9)^2] (* Paolo Xausa, Nov 19 2024 *)
  • PARI
    j=[]; for(n=0,25,j=concat(j,eulerphi( ((10^(n+1)-1)/9)^2) )); j
    
  • PARI
    { for (n=0, 80, a=eulerphi(((10^(n + 1) - 1)/9)^2); write("b063751.txt", n, " ", a) ) } \\ Harry J. Smith, Aug 29 2009

Formula

a(n) = A000042(n) * phi(A000042(n)). - Amiram Eldar, Nov 19 2024

Extensions

Offset corrected by Amiram Eldar, Nov 19 2024

A080150 Let m = Wonderful Demlo number A002477(n); a(n) = square of the sum of digits of m.

Original entry on oeis.org

1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 6724, 7225, 8100, 9409, 11236, 13689, 16900, 21025, 26244, 26569, 27556, 29241, 31684, 34969, 39204, 44521, 51076, 59049, 59536, 61009, 63504, 67081, 71824, 77841, 85264, 94249, 104976, 105625
Offset: 1

Views

Author

Eric W. Weisstein, Jan 31 2003

Keywords

Comments

These numbers are themselves squares, their square roots being in A080151.

Crossrefs

Formula

a(n) = (A080151(n))^2.

Extensions

Name corrected by Robert Israel, Aug 06 2019

A173426 a(n) is obtained by starting with 1, sequentially concatenating all decimal numbers up to n, and then, starting from n-1, sequentially concatenating all decimal numbers down to 1.

Original entry on oeis.org

1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321, 12345678910987654321, 123456789101110987654321, 1234567891011121110987654321, 12345678910111213121110987654321, 123456789101112131413121110987654321
Offset: 1

Views

Author

Umut Uludag, Feb 18 2010

Keywords

Comments

The first prime in this sequence is the 20-digit number a(10) = 12345678910987654321. On Jul 20 2015, Shyam Sunder Gupta reported on the Number Theory Mailing List that he has found what is probably the second prime in the sequence. This is the 2446th term, namely the 17350-digit probable prime 1234567..244524462445..7654321. See A359148. - N. J. A. Sloane, Jul 29 2015 - Aug 03 2015
There are no other (PR)prime members in this sequence for n<60000. - Serge Batalov, Jul 29 2015
David Broadhurst gives heuristic arguments which suggest that this sequence contains infinitely many primes.
See A075023 and A075024 for the smallest and largest prime factor of the terms. - M. F. Hasler, Jul 29 2015
Using summation in decimal length clades, one can obtain analytical expressions for the sequence:
a(n) = A002275(n)^2, for 1 <= n < 10;
a(n) = (120999998998*10^(4*n-28) - 2*10^(2*n-9) + 8790000000121)/99^2, for 10 <= n < 10^2;
a(n) = (120999998998*10^(6*n-227) - (1099022*10^(6*n-406) + 242*10^(3*n-108) - 1087789*10^191)/111^2 + 8790000000121)/99^2, for 10^2 <= n < 10^3; etc. - Serge Batalov, Jul 29 2015
Curiously, 1234567891010987654321 is also a prime (see A259937). - N. J. A. Sloane, Nov 30 2021

References

  • D. Broadhurst, Primes from concatenation: results and heuristics, Number Theory List, Aug 01 2015 and later postings.

Crossrefs

This sequence and A002477 (Wonderful Demlo numbers) agree up to the 9th term.

Programs

  • Maple
    a:= n-> parse(cat($1..n, n-i$i=1..n-1)):
    seq(a(n), n=1..14);  # Alois P. Heinz, Dec 01 2021
  • Mathematica
    Table[FromDigits[Flatten[IntegerDigits/@Join[Range[n],Reverse[Range[ n-1]]]]],{n,15}] (* Harvey P. Dale, Sep 02 2015 *)
  • PARI
    A173426(n)=eval(concat(vector(n*2-1,k,if(kM. F. Hasler, Jul 29 2015
    
  • Python
    def A173426(n): return int(''.join(str(d) for d in range(1,n+1))+''.join(str(d) for d in range(n-1,0,-1))) # Chai Wah Wu, Dec 01 2021

Formula

a(n) = concatenate(1,2,3,...,n-2,n-1,n,n-1,n-2,...,3,2,1).

Extensions

More terms from and minor edits by M. F. Hasler, Jul 29 2015

A075415 Squares of A002280 or numbers (666...6)^2.

Original entry on oeis.org

0, 36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556, 444444444443555555555556, 44444444444435555555555556, 4444444444444355555555555556, 444444444444443555555555555556
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) = 66^2 = 4356.
From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 36 = 9 * 4;
n=2: ................... 4356 = 99 * 44;
n=3: ................. 443556 = 999 * 444;
n=4: ............... 44435556 = 9999 * 4444;
n=5: ............. 4444355556 = 99999 * 44444;
n=6: ........... 444443555556 = 999999 * 444444;
n=7: ......... 44444435555556 = 9999999 * 4444444;
n=8: ....... 4444444355555556 = 99999999 * 44444444;
n=9: ..... 444444443555555556 = 999999999 * 444444444. (End)
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[PadRight[{},n,6]]^2,{n,0,20}] (* or *) LinearRecurrence[ {111,-1110,1000},{0,36,4356},20] (* Harvey P. Dale, May 20 2021 *)

Formula

a(n) = A002280(n)^2 = (6*A002275(n))^2 = 36*A002275(n)^2.
a(n) = (6*(10^n-1)/9)^2 = (4/9)*(10^(2*n) - 2*10^n + 1), which is n-1 4's, followed by a 3, n-1 5's and a 6. - Ignacio Larrosa Cañestro, Feb 26 2005
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = ((A002278(n-1)*10 + 3)*10^(n-1) + A002279(n-1))*10 + 6 for n>0.
a(n) = A002283(n)*A002278(n). (End)
G.f.: 36*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Arkadiusz Wesolowski, Dec 26 2011
From Elmo R. Oliveira, Jul 27 2025: (Start)
E.g.f.: 4*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 36*A002477(n). (End)

Extensions

Edited by Alois P. Heinz, Aug 21 2019 (merged with A102794, submitted by Richard C. Schroeppel, Feb 26 2005)

A059988 a(n) = (10^n - 1)^2.

Original entry on oeis.org

0, 81, 9801, 998001, 99980001, 9999800001, 999998000001, 99999980000001, 9999999800000001, 999999998000000001, 99999999980000000001, 9999999999800000000001, 999999999998000000000001, 99999999999980000000000001, 9999999999999800000000000001, 999999999999998000000000000001
Offset: 0

Views

Author

Henry Bottomley, Mar 07 2001

Keywords

Comments

From James D. Klein, Feb 05 2012: (Start)
The periods of the reciprocals of a(n) are the consecutive integers from 0 to 10^n-1, omitting the one integer 10^n-2, right-justified in field widths of size n.
E.g.:
1/81 = 0.012345679...
1/9801 = 0.000102030405060708091011...9799000102...
1/998001 = 0.000001002003004005...997999000001002... (End)
Sum of first 10^n - 1 odd numbers. - Arkadiusz Wesolowski, Jun 12 2013

Examples

			From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 81 = 9^2;
n=2: ................... 9801 = 99^2;
n=3: ................. 998001 = 999^2;
n=4: ............... 99980001 = 9999^2;
n=5: ............. 9999800001 = 99999^2;
n=6: ........... 999998000001 = 999999^2;
n=7: ......... 99999980000001 = 9999999^2;
n=8: ....... 9999999800000001 = 99999999^2;
n=9: ..... 999999998000000001 = 999999999^2. (End)
		

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.
  • Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 34.

Crossrefs

Programs

Formula

a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.
a(n) = {999... (n times)}^2 = {999... (n times), 000... (n times)} - {999... (n times)}. For example, 999^2 = 999000 - 999 = 998001. - Kyle D. Balliet, Mar 07 2009
a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - Reinhard Zumkeller, May 31 2010
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)
Sum_{n>=1} 1/a(n) = (log(10)*(QPolyGamma(0, 1, 1/10) - log(10/9)) + QPolyGamma(1, 1, 1/10))/log(10)^2 = 0.012448721523422795191... . - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Elmo R. Oliveira, Aug 02 2025
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