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 20 results. Next

A028310 Expansion of (1 - x + x^2) / (1 - x)^2 in powers of x.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71
Offset: 0

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Author

Keywords

Comments

1 followed by the natural numbers.
Molien series for ring of Hamming weight enumerators of self-dual codes (with respect to Euclidean inner product) of length n over GF(4).
Engel expansion of e (see A006784 for definition) [when offset by 1]. - Henry Bottomley, Dec 18 2000
Also the denominators of the series expansion of log(1+x). Numerators are A062157. - Robert G. Wilson v, Aug 14 2015
The right-shifted sequence (with a(0)=0) is an autosequence (of the first kind - see definition in links). - Jean-François Alcover, Mar 14 2017

Examples

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + 8*x^8 + 9*x^9  + ...
		

Crossrefs

Cf. A000007, A000027, A000660 (boustrophedon transform).

Programs

  • Haskell
    a028310 n = 0 ^ n + n
    a028310_list = 1 : [1..]  -- Reinhard Zumkeller, Nov 06 2012
    
  • Magma
    [n eq 0 select 1 else n: n in [0..75]]; // G. C. Greubel, Jan 05 2024
    
  • Maple
    a:= n-> `if`(n=0, 1, n):
    seq(a(n), n=0..60);
  • Mathematica
    Denominator@ CoefficientList[Series[Log[1+x], {x,0,75}], x] (* or *)
    CoefficientList[ Series[(1 -x +x^2)/(1-x)^2, {x,0,75}], x] (* Robert G. Wilson v, Aug 14 2015 *)
    Join[{1}, Range[75]] (* G. C. Greubel, Jan 05 2024 *)
    LinearRecurrence[{2,-1},{1,1,2},80] (* Harvey P. Dale, Jan 29 2025 *)
  • PARI
    {a(n) = (n==0) + max(n, 0)} /* Michael Somos, Feb 02 2004 */
    
  • PARI
    A028310(n)=n+!n  \\ M. F. Hasler, Jan 16 2012
    
  • Python
    def A028310(n): return n|bool(n)^1 # Chai Wah Wu, Jul 13 2023
    
  • SageMath
    [n + int(n==0) for n in range(76)] # G. C. Greubel, Jan 05 2024

Formula

Binomial transform is A005183. - Paul Barry, Jul 21 2003
G.f.: (1 - x + x^2) / (1 - x)^2 = (1 - x^6) /((1 - x) * (1 - x^2) * (1 - x^3)) = (1 + x^3) / ((1 - x) * (1 - x^2)). a(0) = 1, a(n) = n if n>0.
Euler transform of length 6 sequence [ 1, 1, 1, 0, 0, -1]. - Michael Somos Jul 30 2006
G.f.: 1 / (1 - x / (1 - x / (1 + x / (1 - x)))). - Michael Somos, Apr 05 2012
G.f. of A112934(x) = 1 / (1 - a(0)*x / (1 - a(1)*x / ...)). - Michael Somos, Apr 05 2012
a(n) = A000027(n) unless n=0.
a(n) = Sum_{k=0..n} A123110(n,k). - Philippe Deléham, Oct 06 2009
E.g.f: 1+x*exp(x). - Wolfdieter Lang, May 03 2010
a(n) = sqrt(floor[A204503(n+3)/9]). - M. F. Hasler, Jan 16 2012
E.g.f.: 1-x + x*E(0), where E(k) = 2 + x/(2*k+1 - x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
a(n) = A001477(n) + A000007(n). - Miko Labalan, Dec 12 2015 (See the first comment.)

A023110 Squares which remain squares when the last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 16, 49, 169, 256, 361, 1444, 3249, 18496, 64009, 237169, 364816, 519841, 2079364, 4678569, 26666896, 92294449, 341991049, 526060096, 749609641, 2998438564, 6746486769, 38453641216, 133088524969, 493150849009, 758578289296, 1080936581761
Offset: 1

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Author

Keywords

Comments

This A023110 = A031149^2 is the base 10 version of A001541^2 = A055792 (base 2), A001075^2 = A055793 (base 3), A004275^2 = A055808 (base 4), A204520^2 = A055812 (base 5), A204518^2 = A055851 (base 6), A204516^2 = A055859 (base 7), A204514^2 = A055872 (base 8) and A204502^2 = A204503 (base 9). - M. F. Hasler, Sep 28 2014
For the first 4 terms the square has only one digit. It is understood that deleting this digit yields 0. - Colin Barker, Dec 31 2017

References

  • R. K. Guy, Neg and Reg, preprint, Jan 2012.

Crossrefs

Programs

  • Maple
    count:= 1: A[1]:= 0:
    for n from 0 while count < 35 do
      for t in [1,4,6,9] do
        if issqr(10*n^2+t) then
           count:= count+1;
           A[count]:= 10*n^2+t;
        fi
      od
    od:
    seq(A[i],i=1..count); # Robert Israel, Sep 28 2014
  • Mathematica
    fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range@ 1000000, fQ]^2 (* Robert G. Wilson v, Jan 15 2011 *)
  • PARI
    for(n=0,1e7, issquare(n^2\10) & print1(n^2",")) \\  M. F. Hasler, Jan 16 2012

Formula

Appears to satisfy a(n)=1444*a(n-7)+a(n-14)-76*sqrt(a(n-7)*a(n-14)) for n >= 16. For n = 15, 14, 13, ... this would require a(1) = 16, a(0) = 49, a(-1) = 169, ... - Henry Bottomley, May 08 2001; edited by Robert Israel, Sep 28 2014
a(n) = A031149(n)^2. - M. F. Hasler, Sep 28 2014
Conjectures from Colin Barker, Dec 31 2017: (Start)
G.f.: x^2*(1 + 4*x + 9*x^2 + 16*x^3 + 49*x^4 + 169*x^5 + 256*x^6 - 1082*x^7 - 4328*x^8 - 9738*x^9 - 4592*x^10 - 6698*x^11 - 6698*x^12 - 4592*x^13 + 361*x^14 + 1444*x^15 + 3249*x^16 + 256*x^17 + 169*x^18 + 49*x^19 + 16*x^20) / ((1 - x)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)*(1 - 1442*x^7 + x^14)).
a(n) = 1443*a(n-7) - 1443*a(n-14) + a(n-21) for n>22.
(End)

Extensions

More terms from M. F. Hasler, Jan 16 2012

A055793 Numbers k such that k and floor[k/3] are both squares; i.e., squares which remain squares when written in base 3 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 49, 676, 9409, 131044, 1825201, 25421764, 354079489, 4931691076, 68689595569, 956722646884, 13325427460801, 185599261804324, 2585064237799729, 36005300067391876, 501489136705686529, 6984842613812219524, 97286307456665386801, 1355023461779503195684, 18873042157456379352769
Offset: 1

Views

Author

Henry Bottomley, Jul 14 2000

Keywords

Comments

Or, squares of the form 3k^2+1.
See A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

Examples

			a(3) = 49 because 49 = 7^2 = 1211 base 3 and 121 base 3 = 16 = 4^2.
		

Crossrefs

Cf. also A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

Programs

  • Magma
    I:=[0, 1, 4]; [n le 3 select I[n] else 14*Self(n-1) - Self(n-2) - 6: n in [1..30]]; // Vincenzo Librandi, Jan 27 2013
  • Maple
    A055793 := proc(n) coeftayl(x*(1-11*x+4*x^2)/((1-x)*(1-14*x+x^2)), x=0, n); end proc: seq(A055793(n), n=0..20); # Wesley Ivan Hurt, Sep 28 2014
  • Mathematica
    CoefficientList[Series[x*(1 - 11*x + 4*x^2)/((1 - x)*(1 - 14*x + x^2)), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
    LinearRecurrence[{15,-15,1},{0,1,4,49},40] (* Harvey P. Dale, Jun 19 2021 *)
  • PARI
    sq3nsqplus1(n) = { for(x=1,n, y = 3*x*x+1; \ print1(y" ") if(issquare(y),print1(y" ")) ) }
    

Formula

a(n) = 3*A098301(n-2)+1. - R. J. Mathar, Jun 11 2009
a(n) = 14*a(n-1)-a(n-2)-6, with a(0)=1, a(1)=4. (See Brown and Shiue)
a(n) = (A001075(n-2))^2. - Johannes Boot Dec 16 2011, corrected by M. F. Hasler, Jan 15 2012
G.f.: x*(1 - 11*x + 4*x^2)/((1 - x)*(1 - 14*x + x^2)). - M. F. Hasler, Jan 15 2012

Extensions

More terms from Cino Hilliard, Mar 01 2003

A055812 a(n) and floor(a(n)/5) are both squares; i.e., squares which remain squares when written in base 5 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 49, 81, 324, 2209, 15129, 25921, 103684, 710649, 4870849, 8346321, 33385284, 228826129, 1568397609, 2687489281, 10749957124, 73681302249, 505019158609, 865363202001, 3461452808004
Offset: 1

Views

Author

Henry Bottomley, Jul 14 2000

Keywords

Comments

For the first 3 terms, the above "base 5" interpretation is questionable, since they have only 1 digit in base 5. It is understood that dropping this digit yields 0. - M. F. Hasler, Jan 15 2012

Examples

			a(4) = 49 because 49 = 7^2 = 144 base 5 and 14 base 5 = 9 = 3^2.
		

Crossrefs

For analogs in other bases see A055792 (base 2), A055793 (base 3), A055808 (base 4), A055851 (base 6), A204517 (base 7), A204512 (base 8), A204503 (base 9) and A023110 (base 10).
Squares of A204520. The square roots of floor[a(n)/5] are given in A204521.

Programs

  • PARI
    b=5;for(n=1,2e9,issquare(n^2\b) && print1(n^2,","))  \\ M. F. Hasler, Jan 15 2012

Formula

Empirical g.f.: -x^2*(9*x^11 +49*x^10 +324*x^9 +81*x^8 -698*x^7 -698*x^6 -968*x^5 -242*x^4 +49*x^3 +9*x^2 +4*x +1) / ((x -1)*(x +1)*(x^2 -4*x -1)*(x^2 +1)*(x^2 +4*x -1)*(x^4 +18*x^2 +1)). - Colin Barker, Sep 15 2014

Extensions

More terms added and offset changed to 1 by M. F. Hasler, Jan 15 2012

A204502 Numbers such that floor[a(n)^2 / 9] is a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177
Offset: 1

Views

Author

M. F. Hasler, Jan 15 2012

Keywords

Comments

Or, numbers n such that n^2, with its last base-9 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 9.)

Crossrefs

The squares are in A204503, the squares with last base-9 digit dropped in A204504, and the square roots of the latter in A028310.
Cf. A031149=sqrt(A023110) (base 10), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Programs

  • Mathematica
    Select[Range[0,200],IntegerQ[Sqrt[Floor[#^2/9]]]&] (* Harvey P. Dale, May 05 2018 *)
  • PARI
    b=9;for(n=0,200,issquare(n^2\b) & print1(n","))

Formula

Conjecture: a(n) = 3*n-12 for n>5. G.f.: x^2*(x^2+x+1)*(x^3-x+1)/(x-1)^2. [Colin Barker, Nov 23 2012]

A204514 Numbers such that floor(a(n)^2 / 8) is again a square.

Original entry on oeis.org

0, 1, 2, 3, 6, 17, 34, 99, 198, 577, 1154, 3363, 6726, 19601, 39202, 114243, 228486, 665857, 1331714, 3880899, 7761798, 22619537, 45239074, 131836323, 263672646, 768398401, 1536796802, 4478554083, 8957108166, 26102926097, 52205852194, 152139002499, 304278004998, 886731088897
Offset: 1

Views

Author

M. F. Hasler, Jan 15 2012

Keywords

Comments

Or: Numbers whose square, with its last base-8 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 8.)
See A204504 for the squares resulting from truncation of a(n)^2, and A204512 for their square roots. - M. F. Hasler, Sep 28 2014

Crossrefs

Cf. A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Programs

  • Maple
    A204514 := proc(n) coeftayl((x^2+2*x^3-3*x^4-6*x^5)/(1-6*x^2+x^4), x=0, n); end proc: seq(A204514(n), n=1..30); # Wesley Ivan Hurt, Sep 28 2014
  • Mathematica
    CoefficientList[Series[(x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(x (1 - 6*x^2 + x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
    LinearRecurrence[{0,6,0,-1},{0,1,2,3,6},40] (* Harvey P. Dale, Nov 23 2022 *)
  • PARI
    b=8;for(n=0,1e7,issquare(n^2\b) & print1(n","))
    
  • PARI
    A204514(n)=polcoeff((x + 2*x^2 - 3*x^3 - 6*x^4)/(1 - 6*x^2 + x^4+O(x^(n+!n))),n-1,x)

Formula

G.f. = (x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(1 - 6*x^2 + x^4).
a(n) = sqrt(A055872(n)). - M. F. Hasler, Sep 28 2014
a(2n) = A001541(n-1). a(2n+1) = A003499(n-1). - R. J. Mathar, Feb 05 2020

A055851 a(n) and floor(a(n)/6) are both squares; i.e., squares that remain squares when written in base 6 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 25, 100, 729, 2401, 9604, 71289, 235225, 940900, 6985449, 23049601, 92198404, 684502569, 2258625625, 9034502500, 67074266169, 221322261601, 885289046404, 6572593581849, 21687323011225, 86749292044900
Offset: 1

Views

Author

Henry Bottomley, Jul 14 2000

Keywords

Comments

For the first 3 terms, the above "base 6" interpretation is questionable, since they have only 1 digit in base 6. It is understood that dropping this digit yields 0. - M. F. Hasler, Jan 15 2012
Base-6 analog of A055792 (base 2), A055793 (base 3), A055808 (base 4), A055812 (base 5), A204517 (base 7), A204503 (base 9) and A023110 (base 10). - M. F. Hasler, Jan 15 2012

Examples

			a(5) = 100 because 100 = 10^2 = 244 base 6 and 24 base 6 = 16 = 4^2.
		

Crossrefs

Cf. A023110.

Programs

  • PARI
    b=6;for(n=1,2e9,issquare(n^2\b) & print1(n^2,",")) \\ M. F. Hasler, Jan 15 2012

Formula

a(n) = A204518(n)^2. - M. F. Hasler, Jan 15 2012
Empirical g.f.: -x^2*(9*x^8+100*x^7+25*x^6-162*x^5-296*x^4-74*x^3+9*x^2+4*x+1) / ((x-1)*(x^2+x+1)*(x^6-98*x^3+1)). - Colin Barker, Sep 15 2014

Extensions

More terms added and offset changed to 1 by M. F. Hasler, Jan 16 2012

A055872 a(n) and floor(a(n)/8) are both squares; i.e., squares that remain squares when written in base 8 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 36, 289, 1156, 9801, 39204, 332929, 1331716, 11309769, 45239076, 384199201, 1536796804, 13051463049, 52205852196, 443365544449, 1773462177796, 15061377048201, 60245508192804
Offset: 1

Views

Author

Henry Bottomley, Jul 14 2000

Keywords

Comments

For the first 3 terms which have only 1 digit in base 8, removing this digit is meant to yield 0.
Base-8 analog of A055792 (base 2), A055793 (base 3), A055808 (base 4), A055812 (base 5), A055851 (base 6), A055859 (base 7), A204503 (base 9) and A023110 (base 10). - M. F. Hasler, Jan 15 2012

Examples

			a(5) = 289 because 289 = 17^2 = 441 base 8 and 44 base 8 = 36 = 6^2.
		

Crossrefs

Cf. A023110, A055792 (bisection).

Programs

  • Mathematica
    Select[Range[0,8*10^6]^2,IntegerQ[Sqrt[FromDigits[Most[ IntegerDigits[ #,8]], 8]]]&] (* Harvey P. Dale, Aug 02 2016 *)
  • PARI
    b=8;for(n=1,200,issquare(n^2\b) && print1(n^2,",")) \\ M. F. Hasler, Jan 15 2012

Formula

a(n) = A204514(n)^2. - M. F. Hasler, Jan 15 2012
Empirical g.f.: -x^2*(4*x+1)*(9*x^4-26*x^2+1) / ((x-1)*(x+1)*(x^2-6*x+1)*(x^2+6*x+1)). - Colin Barker, Sep 15 2014

Extensions

More terms added and offset changed to 1 by M. F. Hasler, Jan 15 2012

A204517 Square root of floor[A055859(n)/7].

Original entry on oeis.org

0, 0, 0, 1, 3, 6, 17, 48, 96, 271, 765, 1530, 4319, 12192, 24384, 68833, 194307, 388614, 1097009, 3096720, 6193440, 17483311, 49353213, 98706426, 278635967, 786554688, 1573109376, 4440692161, 12535521795, 25071043590, 70772438609, 199781794032, 399563588064
Offset: 1

Views

Author

M. F. Hasler, Jan 15 2012

Keywords

Crossrefs

See also A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Programs

  • PARI
    b=7;for(n=1,2e9,issquare(n^2\b) & print1(sqrtint(n^2\b),","))
    
  • PARI
    A204517(n)=polcoeff((x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6+O(x^n)),n)

Formula

A204517(n) = sqrt(floor(A204516(n)^2/7)).
G.f. = (x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6)

A204512 Square roots of [A055872/8]: Their square written in base 8, with some digit appended, is again a square.

Original entry on oeis.org

0, 0, 0, 1, 2, 6, 12, 35, 70, 204, 408, 1189, 2378, 6930, 13860, 40391, 80782, 235416, 470832, 1372105, 2744210, 7997214, 15994428, 46611179, 93222358, 271669860, 543339720, 1583407981, 3166815962, 9228778026, 18457556052, 53789260175, 107578520350
Offset: 1

Views

Author

M. F. Hasler, Jan 15 2012

Keywords

Comments

Base-8 analog of A031150. The square of the terms (= truncated squares A055872) are listed in A204504.

Crossrefs

See also A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Programs

  • Mathematica
    CoefficientList[Series[(x^4 (1+2x))/(1-6x^2+x^4),{x,0,40}],x] (* Harvey P. Dale, Nov 30 2020 *)
  • PARI
    b=8;for(n=1,1e7,issquare(n^2\b) & print1(sqrtint(n^2\b)","))
    
  • PARI
    a(n)=polcoeff((2*x^5 + x^4)/(x^4 - 6*x^2 + 1+O(x^n)),n)

Formula

G.f. = x^4*(1 + 2*x)/(1 - 6*x^2 + x^4)
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