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 37 results. Next

A144756 Partial products of successive terms of A017101; a(0)=1 .

Original entry on oeis.org

1, 3, 33, 627, 16929, 592515, 25478145, 1299385395, 76663738305, 5136470466435, 385235284982625, 31974528653557875, 2909682107473766625, 288058528639902895875, 30822262564469609858625, 3544560194914005133741875, 435980903974422631450250625, 57113498420649364719982831875
Offset: 0

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Author

Philippe Deléham, Sep 20 2008

Keywords

Examples

			a(0)=1, a(1)=3, a(2)=3*11=33, a(3)=3*11*19=627, a(4)=3*11*19*27=16929, ...
		

Crossrefs

Programs

  • Mathematica
    Join[{1},FoldList[Times,8Range[0,20]+3]] (* Harvey P. Dale, Aug 11 2019 *)

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*3^k*8^(n-k).
a(n) = (-5)^n*sum_{k=0..n} (8/5)^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 2/G(0), where G(k)= 1 + 1/(1 - 2*x*(8*k+3)/(2*x*(8*k+3) - 1 + 16*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
a(n) +(-8*n+5)*a(n-1)=0. - R. J. Mathar, Sep 04 2016
From Ilya Gutkovskiy, Mar 23 2017: (Start)
E.g.f.: 1/(1 - 8*x)^(3/8).
a(n) ~ sqrt(2*Pi)*8^n*n^n/(exp(n)*n^(1/8)*Gamma(3/8)). (End)
Sum_{n>=0} 1/a(n) = 1 + (e/8^5)^(1/8)*(Gamma(3/8) - Gamma(3/8, 1/8)). - Amiram Eldar, Dec 20 2022

Extensions

a(11) corrected by Ilya Gutkovskiy, Mar 23 2017

A348845 Part two of the trisection of A017101: a(n) = 11 + 24*n.

Original entry on oeis.org

11, 35, 59, 83, 107, 131, 155, 179, 203, 227, 251, 275, 299, 323, 347, 371, 395, 419, 443, 467, 491, 515, 539, 563, 587, 611, 635, 659, 683, 707, 731, 755, 779, 803, 827, 851, 875, 899, 923, 947, 971, 995, 1019, 1043, 1067
Offset: 0

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Author

Wolfdieter Lang, Dec 11 2021

Keywords

Comments

The trisection of A017101 = {3 + 8*k}A017077%20=%20%7B3*(1%20+%2012*n)%7D">{k>=0} gives 3*A017077 = {3*(1 + 12*n)}{n>=0}, {a(n)}A350051%20=%20%7B19%20+%2024*n%7D">{n >= 0} and A350051 = {19 + 24*n}{n>=0}. These three sequences are congruent to 3 modulo 8 and to 3, 5, and 1 modulo 6, respectively.

Crossrefs

Programs

  • Mathematica
    24 * Range[0, 44] + 11 (* Amiram Eldar, Dec 18 2021 *)

Formula

a(n) = 11 + 24*n = 11 + A008606(n), for n >= 0
a(n) = 2*a(n-1) - a(n-2), for n >= 1, with a(-1) = -13, a(0) = 11.
G.f.: (11 + 13*x)/(1-x)^2.
E.g.f.: (11 + 24*x)*exp(x).

A350051 Part three of the trisection of A017101: a(n) = 19 + 24*n.

Original entry on oeis.org

19, 43, 67, 91, 115, 139, 163, 187, 211, 235, 259, 283, 307, 331, 355, 379, 403, 427, 451, 475, 499, 523, 547, 571, 595, 619, 643, 667, 691, 715, 739, 763, 787, 811, 835, 859, 883, 907, 931, 955, 979, 1003, 1027, 1051, 1075
Offset: 0

Views

Author

Wolfdieter Lang, Dec 11 2021

Keywords

Comments

The trisection of A017101 = {3 + 8*k}A017077%20=%20%7B3*(1%20+%2012*n)%7D">{k>=0} gives 3*A017077 = {3*(1 + 12*n)}{n>=0}, {A348845(n)}{n >= 0} and {a(n)}{n>=0}. These three sequences are congruent to 3 modulo 8 and to 3, 5, and 1 modulo 6, respectively.

Crossrefs

Programs

Formula

a(n) = 19 + 24*n = 19 + A008606(n), for n >= 0
a(n) = 2*a(n-1) - a(n-2), for n >= 1, with a(-1) = -5, a(0) = 19.
G.f.: (19 + 5*x)/(1-x)^2.
E.g.f.: (19 + 24*x)*exp(x).

A000069 Odious numbers: numbers with an odd number of 1's in their binary expansion.

Original entry on oeis.org

1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59, 61, 62, 64, 67, 69, 70, 73, 74, 76, 79, 81, 82, 84, 87, 88, 91, 93, 94, 97, 98, 100, 103, 104, 107, 109, 110, 112, 115, 117, 118, 121, 122, 124, 127, 128
Offset: 1

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Author

Keywords

Comments

This sequence and A001969 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.
In French: les nombres impies.
Has asymptotic density 1/2, since exactly 2 of the 4 numbers 4k, 4k+1, 4k+2, 4k+3 have an even sum of bits, while the other 2 have an odd sum. - Jeffrey Shallit, Jun 04 2002
Nim-values for game of mock turtles played with n coins.
A115384(n) = number of odious numbers <= n; A000120(a(n)) = A132680(n). - Reinhard Zumkeller, Aug 26 2007
Indices of 1's in the Thue-Morse sequence A010060. - Tanya Khovanova, Dec 29 2008
For any positive integer m, the partition of the set of the first 2^m positive integers into evil ones E and odious ones O is a fair division for any polynomial sequence p(k) of degree less than m, that is, Sum_{k in E} p(k) = Sum_{k in O} p(k) holds for any polynomial p with deg(p) < m. - Pietro Majer, Mar 15 2009
For n>1 let b(n) = a(n-1). Then b(b(n)) = 2b(n). - Benoit Cloitre, Oct 07 2010
Lexicographically earliest sequence of distinct nonnegative integers with no term being the binary exclusive OR of any terms. The equivalent sequence for addition or for subtraction is A005408 (the odd numbers) and for multiplication is A026424. - Peter Munn, Jan 14 2018
Numbers of the form m XOR (2*m+1) for some m >= 0. - Rémy Sigrist, Apr 14 2022

Examples

			For k=2, x=0 and x=0.2 we respectively have 1^2 + 2^2 + 4^2 + 7^2 = 0^2 + 3^2 + 5^2 + 6^2 = 70;
(1.2)^2 + (2.2)^2 + (4.2)^2 + (7.2)^2 = (0.2)^2 + (3.2)^2 + (5.2)^2 + (6.2)^2 = 75.76;
for k=3, x=1.8, we have (2.8)^3 + (3.8)^3 + (5.8)^3 + (8.8)^3 + (9.8)^3 + (12.8)^3 + (14.8)^3 + (15.8)^3 = (1.8)^3 + (4.8)^3 + (6.8)^3 + (7.8)^3 + (10.8)^3 + (11.8)^3 + (13.8)^3 + (16.8)^3 = 11177.856. - _Vladimir Shevelev_, Jan 16 2012
		

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 433.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 22.
  • Vladimir S. Shevelev, On some identities connected with the partition of the positive integers with respect to the Morse sequence, Izv. Vuzov of the North-Caucasus region, Nature sciences 4 (1997), 21-23 (in Russian).
  • N. J. A. Sloane, A handbook of Integer Sequences, Academic Press, 1973 (including this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
Complement of A001969 (the evil numbers). Cf. A133009.
a(n) = 2*n + 1 - A010060(n) = A001969(n) + (-1)^A010060(n).
First differences give A007413.
Note that A000079, A083420, A002042, A002089, A132679 are subsequences.
See A027697 for primes, also A230095.
Cf. A005408 (odd numbers), A006068, A026424.

Programs

  • Haskell
    a000069 n = a000069_list !! (n-1)
    a000069_list = [x | x <- [0..], odd $ a000120 x]
    -- Reinhard Zumkeller, Feb 01 2012
    
  • Magma
    [ n: n in [1..130] | IsOdd(&+Intseq(n, 2)) ]; // Klaus Brockhaus, Oct 07 2010
    
  • Maple
    s := proc(n) local i,j,k,b,sum,ans; ans := [ ]; j := 0; for i while jA000069 := n->t1[n]; # s(k) gives first k terms.
    is_A000069 := n -> type(add(i,i=convert(n,base,2)),odd):
    seq(`if`(is_A000069(i),i,NULL),i=0..40); # Peter Luschny, Feb 03 2011
  • Mathematica
    Select[Range[300], OddQ[DigitCount[ #, 2][[1]]] &] (* Stefan Steinerberger, Mar 31 2006 *)
    a[ n_] := If[ n < 1, 0, 2 n - 1 - Mod[ Total @ IntegerDigits[ n - 1, 2], 2]]; (* Michael Somos, Jun 01 2013 *)
  • PARI
    {a(n) = if( n<1, 0, 2*n - 1 - subst( Pol(binary( n-1)), x, 1) % 2)}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    {a(n) = if( n<2, n==1, if( n%2, a((n+1)/2) + n-1, -a(n/2) + 3*(n-1)))}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    a(n)=2*n-1-hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    [n for n in range(1, 201) if bin(n)[2:].count("1") % 2] # Indranil Ghosh, May 03 2017
    
  • Python
    def A000069(n): return ((m:=n-1)<<1)+(m.bit_count()&1^1) # Chai Wah Wu, Mar 03 2023

Formula

G.f.: 1 + Sum_{k>=0} (t*(2+2t+5t^2-t^4)/(1-t^2)^2) * Product_{j=0..k-1} (1-x^(2^j)), t=x^2^k. - Ralf Stephan, Mar 25 2004
a(n+1) = (1/2) * (4*n + 1 + (-1)^A000120(n)). - Ralf Stephan, Sep 14 2003
Numbers n such that A010060(n) = 1. - Benoit Cloitre, Nov 15 2003
a(2*n+1) + a(2*n) = A017101(n) = 8*n+3. a(2*n+1) - a(2*n) gives the Thue-Morse sequence (1, 3 version): 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, ... A001969(n) + A000069(n) = A016813(n) = 4*n+1. - Philippe Deléham, Feb 04 2004
(-1)^a(n) = 2*A010060(n)-1. - Benoit Cloitre, Mar 08 2004
a(1) = 1; for n > 1: a(2*n) = 6*n-3 -a(n), a(2*n+1) = a(n+1) + 2*n. - Corrected by Vladimir Shevelev, Sep 25 2011
For k >= 1 and for every real (or complex) x, we have Sum_{i=1..2^k} (a(i)+x)^s = Sum_{i=1..2^k} (A001969(i)+x)^s, s=0..k.
For x=0, s <= k-1, this is known as Prouhet theorem (see J.-P. Allouche and Jeffrey Shallit, The Ubiquitous Prouhet-Thue-Morse Sequence). - Vladimir Shevelev, Jan 16 2012
a(n+1) mod 2 = 1 - A010060(n) = A010059(n). - Robert G. Wilson v, Jan 18 2012
A005590(a(n)) > 0. - Reinhard Zumkeller, Apr 11 2012
A106400(a(n)) = -1. - Reinhard Zumkeller, Apr 29 2012
a(n+1) = A006068(n) XOR (2*A006068(n) + 1). - Rémy Sigrist, Apr 14 2022

A001969 Evil numbers: nonnegative integers with an even number of 1's in their binary expansion.

Original entry on oeis.org

0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58, 60, 63, 65, 66, 68, 71, 72, 75, 77, 78, 80, 83, 85, 86, 89, 90, 92, 95, 96, 99, 101, 102, 105, 106, 108, 111, 113, 114, 116, 119, 120, 123, 125, 126, 129
Offset: 1

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Author

Keywords

Comments

This sequence and A000069 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.
In French: les nombres païens.
Theorem: First differences give A036585. (Observed by Franklin T. Adams-Watters.)
Proof from Max Alekseyev, Aug 30 2006 (edited by N. J. A. Sloane, Jan 05 2021): (Start)
Observe that if the last bit of a(n) is deleted, we get the nonnegative numbers 0, 1, 2, 3, ... in order.
The last bit in a(n+1) is 1 iff the number of bits in n is odd, that is, iff A010060(n+1) is 1.
So, taking into account the different offsets here and in A010060, we have a(n) = 2*(n-1) + A010060(n-1).
Therefore the first differences of the present sequence equal 2 + first differences of A010060, which equals A036585. QED (End)
Integers k such that A010060(k-1)=0. - Benoit Cloitre, Nov 15 2003
Indices of zeros in the Thue-Morse sequence A010060 shifted by 1. - Tanya Khovanova, Feb 13 2009
Conjecture, checked up to 10^6: a(n) is also the sequence of numbers k representable as k = ror(x) XOR rol(x) (for some integer x) where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 14 2016
From Charlie Neder, Oct 07 2018: (Start)
Conjecture is true: ror(x) and rol(x) have an even number of 1 bits in total (= 2 * A000120(x)), and XOR preserves the parity of this total, so the resulting number must have an even number of 1 bits. An x can be constructed corresponding to a(n) like so:
If the number of bits in a(n) is even, add a leading 0 so a(n) is 2k+1 bits long.
Do an inverse shuffle on a(n), then "divide" by 11, rotate the result k bits to the right, and shuffle to get x. (End)
Numbers of the form m XOR (2*m) for some m >= 0. - Rémy Sigrist, Feb 07 2021
The terms "evil numbers" and "odious numbers" were coined by Richard K. Guy, c. 1976 (Haque and Shallit, 2016) and appeared in the book by Berlekamp et al. (Vol. 1, 1st ed., 1982). - Amiram Eldar, Jun 08 2021

References

  • Elwyn R. Berlekamp, John H. Conway, Richard K. Guy, Winning Ways for Your Mathematical Plays, Volume 1, 2nd ed., A K Peters, 2001, chapter 14, p. 110.
  • Hugh L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 208.
  • Donald J. Newman, A Problem Seminar, Springer; see Problem #89.
  • Vladimir S. Shevelev, On some identities connected with the partition of the positive integers with respect to the Morse sequence, Izv. Vuzov of the North-Caucasus region, Nature sciences 4 (1997), 21-23 (Russian).
  • 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).

Crossrefs

Complement of A000069 (the odious numbers). Cf. A133009.
a(n)=2*n+A010060(n)=A000069(n)-(-1)^A010060(n). Cf. A018900.
The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
Cf. A036585 (differences), A010060, A006364.
For primes see A027699, also A130593.

Programs

  • Haskell
    a001969 n = a001969_list !! (n-1)
    a001969_list = [x | x <- [0..], even $ a000120 x]
    -- Reinhard Zumkeller, Feb 01 2012
    
  • Magma
    [ n : n in [0..129] | IsEven(&+Intseq(n,2)) ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
    
  • Maple
    s := proc(n) local i,j,ans; ans := [ ]; j := 0; for i from 0 while jA001969 := n->t1[n]; # s(k) gives first k terms.
    # Alternative:
    seq(`if`(add(k, k=convert(n,base,2))::even, n, NULL), n=0..129); # Peter Luschny, Jan 15 2021
    # alternative for use outside this sequence
    isA001969 := proc(n)
        add(d,d=convert(n,base,2)) ;
        type(%,'even') ;
    end proc:
    A001969 := proc(n)
        option remember ;
        local a;
        if n = 0 then
            1;
        else
            for a from procname(n-1)+1 do
                if isA001969(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A001969(n),n=1..200) ; # R. J. Mathar, Aug 07 2022
  • Mathematica
    Select[Range[0,300], EvenQ[DigitCount[ #, 2][[1]]] &]
    a[ n_] := If[ n < 1, 0, With[{m = n - 1}, 2 m + Mod[-Total@IntegerDigits[m, 2], 2]]]; (* Michael Somos, Jun 09 2019 *)
  • PARI
    a(n)=n-=1; 2*n+subst(Pol(binary(n)),x,1)%2
    
  • PARI
    a(n)=if(n<1,0,if(n%2==0,a(n/2)+n,-a((n-1)/2)+3*n))
    
  • PARI
    a(n)=2*(n-1)+hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    def ok(n): return bin(n)[2:].count('1') % 2 == 0
    print(list(filter(ok, range(130)))) # Michael S. Branicky, Jun 02 2021
    
  • Python
    from itertools import chain, count, islice
    def A001969_gen(): # generator of terms
        return chain((0,),chain.from_iterable((sorted(n^ n<<1 for n in range(2**l,2**(l+1))) for l in count(0))))
    A001969_list = list(islice(A001969_gen(),30)) # Chai Wah Wu, Jun 29 2022
    
  • Python
    def A001969(n): return ((m:=n-1).bit_count()&1)+(m<<1) # Chai Wah Wu, Mar 03 2023

Formula

a(n+1) - A001285(n) = 2n-1 has been verified for n <= 400. - John W. Layman, May 16 2003 [This can be directly verified by comparing Ralf Stephan's formulas for this sequence (see below) and for A001285. - Jianing Song, Nov 04 2024]
Note that 2n+1 is in the sequence iff 2n is not and so this sequence has asymptotic density 1/2. - Franklin T. Adams-Watters, Aug 23 2006
a(n) = (1/2) * (4n - 3 - (-1)^A000120(n-1)). - Ralf Stephan, Sep 14 2003
G.f.: Sum_{k>=0} (t(3+2t+3t^2)/(1-t^2)^2) * Product_{l=0..k-1} (1-x^(2^l)), where t = x^2^k. - Ralf Stephan, Mar 25 2004
a(2*n+1) + a(2*n) = A017101(n-1) = 8*n-5.
a(2*n) - a(2*n-1) gives the Thue-Morse sequence (3, 1 version): 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, .... A001969(n) + A000069(n) = A016813(n-1) = 4*n-3. - Philippe Deléham, Feb 04 2004
a(1) = 0; for n > 1: a(n) = 3*n-3 - a(n/2) if n even, a(n) = a((n+1)/2)+n-1 if n odd.
Let b(n) = 1 if sum of digits of n is even, -1 if it is odd; then Shallit (1985) showed that Product_{n>=0} ((2n+1)/(2n+2))^b(n) = 1/sqrt(2).
a(n) = 2n - 2 + A010060(n-1). - Franklin T. Adams-Watters, Aug 28 2006
A005590(a(n-1)) <= 0. - Reinhard Zumkeller, Apr 11 2012
A106400(a(n-1)) = 1. - Reinhard Zumkeller, Apr 29 2012
a(n) = (a(n-1) + 2) XOR A010060(a(n-1) + 2). - Falk Hüffner, Jan 21 2022
a(n+1) = A006068(n) XOR (2*A006068(n)). - Rémy Sigrist, Apr 14 2022

Extensions

More terms from Robin Trew (trew(AT)hcs.harvard.edu)

A004767 a(n) = 4*n + 3.

Original entry on oeis.org

3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135, 139, 143, 147, 151, 155, 159, 163, 167, 171, 175, 179, 183, 187, 191, 195, 199, 203, 207, 211, 215, 219, 223
Offset: 0

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Author

Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(12).
Binary expansion ends 11.
These the numbers for which zeta(2*x+1) needs just 2 terms to be evaluated. - Jorge Coveiro, Dec 16 2004 [This comment needs clarification]
a(n) is the smallest k such that for every r from 0 to 2n - 1 there exist j and i, k >= j > i > 2n - 1, such that j - i == r (mod (2n - 1)), with (k, (2n - 1)) = (j,(2n - 1)) = (i, (2n - 1)) = 1. - Amarnath Murthy, Sep 24 2003
Complement of A004773. - Reinhard Zumkeller, Aug 29 2005
Any (4n+3)-dimensional manifold endowed with a mixed 3-Sasakian structure is an Einstein space with Einstein constant lambda = 4n + 2 [Theorem 3, p. 10 of Ianus et al.]. - Jonathan Vos Post, Nov 24 2008
Solutions to the equation x^(2*x) = 3*x (mod 4*x). - Farideh Firoozbakht, May 02 2010
Subsequence of A022544. - Vincenzo Librandi, Nov 20 2010
First differences of A084849. - Reinhard Zumkeller, Apr 02 2011
Numbers n such that {1, 2, 3, ..., n} is a losing position in the game of Nim. - Franklin T. Adams-Watters, Jul 16 2011
Numbers n such that there are no primes p that satisfy the relationship p XOR n = p + n. - Brad Clardy, Jul 22 2012
The XOR of all numbers from 1 to a(n) is 0. - David W. Wilson, Apr 21 2013
A089911(4*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
First differences of A014105. - Ivan N. Ianakiev, Sep 21 2013
All triangular numbers in the sequence are congruent to {3, 7} mod 8. - Ivan N. Ianakiev, Nov 12 2013
Apart from the initial term, length of minimal path on an n-dimensional cubic lattice (n > 1) of side length 2, until a self-avoiding walk gets stuck. Construct a path connecting all 2n points orthogonally adjacent from the center, ending at the center. Starting at any point adjacent to the center, there are 2 steps to reach each of the remaining 2n - 1 points, resulting in path length 4n - 2 with a final step connecting the center, for a total path length of 4n - 1, comprising 4n points. - Matthew Lehman, Dec 10 2013
a(n-1), n >= 1, appears as first column in the triangles A238476 and A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014
For the Collatz Conjecture, we identify two types of odd numbers. This sequence contains all the ascenders: where (3*a(n) + 1) / 2 is odd and greater than a(n). See A016813 for the descenders. - Jaroslav Krizek, Jul 29 2016

Examples

			G.f. = 3 + 7*x + 11*x^2 + 15*x^3 + 19*x^4 + 23*x^5 + 27*x^6 + 31*x^7 + ...
		

References

  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 85.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999. See Theorem 8.1 on page 240.

Crossrefs

Cf. A017101 and A004771 (bisection: 3 and 7 mod 8).
Cf. A016838 (square).

Programs

Formula

G.f.: (3+x)/(1-x)^2. - Paul Barry, Feb 27 2003
a(n) = 2*a(n-1) - a(n-2) for n > 1, a(0) = 3, a(1) = 7. - Philippe Deléham, Nov 03 2008
a(n) = A017137(n)/2. - Reinhard Zumkeller, Jul 13 2010
a(n) = 8*n - a(n-1) + 2 for n > 0, a(0) = 3. - Vincenzo Librandi, Nov 20 2010
a(n) = A005408(A005408(n)). - Reinhard Zumkeller, Jun 27 2011
a(n) = 3 + A008586(n). - Omar E. Pol, Jul 27 2012
a(n) = A014105(n+1) - A014105(n). - Michel Marcus, Sep 21 2013
a(n) = A016813(n) + 2. - Jean-Bernard François, Sep 27 2013
a(n) = 4*n - 1, with offset 1. - Wesley Ivan Hurt, Mar 12 2014
From Ilya Gutkovskiy, Jul 29 2016: (Start)
E.g.f.: (3 + 4*x)*exp(x).
Sum_{n >= 0} (-1)^n/a(n) = (Pi + 2*log(sqrt(2) - 1))/(4*sqrt(2)) = A181049. (End)

A004771 a(n) = 8*n + 7. Or, numbers whose binary expansion ends in 111.

Original entry on oeis.org

7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 119, 127, 135, 143, 151, 159, 167, 175, 183, 191, 199, 207, 215, 223, 231, 239, 247, 255, 263, 271, 279, 287, 295, 303, 311, 319, 327, 335, 343, 351, 359, 367, 375, 383, 391, 399, 407, 415, 423, 431
Offset: 0

Views

Author

Keywords

Comments

These numbers cannot be expressed as the sum of 3 squares. - Artur Jasinski, Nov 22 2006
These numbers cannot be perfect squares. - Cino Hilliard, Sep 03 2006
a(n-2), n >= 2, appears in the second column of triangle A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014
The initial terms 7, 15, 23, 31 are the generating set for the rest of the sequence in the sense that, by Lagrange's Four Square Theorem, any number n of the form 8*k+7 can always be written as a sum of no fewer than four squares, and if n = a^2 + b^2 + c^2 + d^2, then (a mod 4)^2 + (b mod 4)^2 + (c mod 4)^2 + (d mod 4)^2 must be one of 7, 15, 23, 31. - Walter Kehowski, Jul 07 2014
Define a set of consecutive positive odd numbers {1, 3, 5, ..., 12*n + 9} and skip the number 6*n + 5. Then the contraharmonic mean of that set gives this sequence. For example, ContraharmonicMean[{1, 3, 7, 9}] = 7. - Hilko Koning, Aug 27 2018
Jacobi symbol (2, a(n)) = Kronecker symbol (a(n), 2) = 1. - Jianing Song, Aug 28 2018

References

  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 246.

Crossrefs

Cf. A007522 (primes), subsequence of A047522.

Programs

Formula

O.g.f: (7 + x)/(1 - x)^2 = 8/(1 - x)^2 - 1/(1 - x). - R. J. Mathar, Nov 30 2007
a(n) = 2*a(n-1) - a(n-2) for n >= 2. - Vincenzo Librandi, May 28 2011
A056753(a(n)) = 7. - Reinhard Zumkeller, Aug 23 2009
a(n) = t(t(t(n))), where t(i) = 2*i + 1.
a(n) = A004767(2*n+1), for n >= 0. See also A004767(2*n) = A017101(n). - Wolfdieter Lang, Feb 03 2022
From Elmo R. Oliveira, Apr 11 2024: (Start)
E.g.f.: exp(x)*(7 + 8*x).
a(n) = A033954(n+1) - A033954(n). (End)

A080851 Square array of pyramidal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 3, 1, 4, 6, 1, 5, 10, 10, 1, 6, 14, 20, 15, 1, 7, 18, 30, 35, 21, 1, 8, 22, 40, 55, 56, 28, 1, 9, 26, 50, 75, 91, 84, 36, 1, 10, 30, 60, 95, 126, 140, 120, 45, 1, 11, 34, 70, 115, 161, 196, 204, 165, 55, 1, 12, 38, 80, 135, 196, 252, 288, 285, 220, 66, 1, 13, 42, 90, 155, 231, 308, 372, 405, 385, 286, 78
Offset: 0

Views

Author

Paul Barry, Feb 21 2003

Keywords

Comments

The first row contains the triangular numbers, which are really two-dimensional, but can be regarded as degenerate pyramidal numbers. - N. J. A. Sloane, Aug 28 2015

Examples

			Array begins (n>=0, k>=0):
1,  3,  6, 10,  15,  21,  28,  36,  45,   55, ... A000217
1,  4, 10, 20,  35,  56,  84, 120, 165,  220, ... A000292
1,  5, 14, 30,  55,  91, 140, 204, 285,  385, ... A000330
1,  6, 18, 40,  75, 126, 196, 288, 405,  550, ... A002411
1,  7, 22, 50,  95, 161, 252, 372, 525,  715, ... A002412
1,  8, 26, 60, 115, 196, 308, 456, 645,  880, ... A002413
1,  9, 30, 70, 135, 231, 364, 540, 765, 1045, ... A002414
1, 10, 34, 80, 155, 266, 420, 624, 885, 1210, ... A007584
		

Crossrefs

Numerous sequences in the database are to be found in the array. Rows include the pyramidal numbers A000217, A000292, A000330, A002411, A002412, A002413, A002414, A007584, A007585, A007586.
Columns include or are closely related to A017029, A017113, A017017, A017101, A016777, A017305. Diagonals include A006325, A006484, A002417.
Cf. A057145, A027660 (antidiagonal sums).
See A257199 for another version of this array.

Programs

  • Derive
    vector(vector(poly_coeff(Taylor((1+kx)/(1-x)^4,x,11),x,n),n,0,11),k,-1,10) VECTOR(VECTOR(comb(k+2,2)+comb(k+2,3)n, k, 0, 11), n, 0, 11)
  • Maple
    A080851 := proc(n,k)
        binomial(k+3,3)+(n-1)*binomial(k+2,3) ;
    end proc:
    seq( seq(A080851(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021
  • Mathematica
    pyramidalFigurative[ ngon_, rank_] := (3 rank^2 + rank^3 (ngon - 2) - rank (ngon - 5))/6; Table[ pyramidalFigurative[n-k-1, k], {n, 4, 15}, {k, n-3}] // Flatten (* Robert G. Wilson v, Sep 15 2015 *)

Formula

T(n, k) = binomial(k+3, 3) + (n-1)*binomial(k+2, 3), corrected Oct 01 2021.
T(n, k) = T(n-1, k) + C(k+2, 3) = T(n-1, k) + k*(k+1)*(k+2)/6.
G.f. for rows: (1 + n*x)/(1-x)^4, n>=-1.
T(n,k) = sum_{j=1..k+1} A057145(n+2,j). - R. J. Mathar, Jul 28 2016

A017137 a(n) = 8*n + 6.

Original entry on oeis.org

6, 14, 22, 30, 38, 46, 54, 62, 70, 78, 86, 94, 102, 110, 118, 126, 134, 142, 150, 158, 166, 174, 182, 190, 198, 206, 214, 222, 230, 238, 246, 254, 262, 270, 278, 286, 294, 302, 310, 318, 326, 334, 342, 350, 358, 366, 374, 382, 390, 398, 406, 414, 422, 430
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

First differences of A002943. - Aaron David Fairbanks, May 13 2014

Examples

			G.f. = 6 + 14*x + 22*x^2 + 30*x^3 + 38*x^4 + 46*x^5 + 54*x^6 + 62*x^7 + ...
		

Crossrefs

Programs

Formula

a(n) = 2*A004767(n) = A000290(A017245(n)) - A156676(n+1). - Reinhard Zumkeller, Jul 13 2010
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Jun 07 2011
A089911(3*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
From Michael Somos, May 15 2014: (Start)
G.f.: (6 + 2*x)/(1 - x)^2.
E.g.f.: (6 + 8*x)*exp(x). (End)
Sum_{n>=0} (-1)^n/a(n) = (Pi + log(3-2*sqrt(2)))/(8*sqrt(2)). - Amiram Eldar, Dec 11 2021
a(n) = A016825(2*n+1). - Elmo R. Oliveira, Apr 12 2025

A004770 Numbers of the form 8k+5; or, numbers whose binary expansion ends in 101.

Original entry on oeis.org

5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, 101, 109, 117, 125, 133, 141, 149, 157, 165, 173, 181, 189, 197, 205, 213, 221, 229, 237, 245, 253, 261, 269, 277, 285, 293, 301, 309, 317, 325, 333, 341, 349, 357, 365, 373, 381, 389, 397, 405, 413, 421, 429, 437, 445
Offset: 1

Views

Author

Keywords

Comments

Only numbers of the form 8k+5 may be written as a sum of 5 odd squares. Examples: 5 = 1+1+1+1+1, 13 = 9+1+1+1+1, 21 = 9+9+1+1+1, 29 = 25+1+1+1+1= 9+9+9+1+1, 37 = 9+9+9+9+1 = 25+9+1+1+1, 45 = 25+9+9+1+1=9+9+9+9+9, 53 = 49+1+1+1+1 = 25+25+1+1+1 = 25+9+9+9+1, ... - Philippe Deléham, Sep 03 2005
Positive solutions to the equation x == 5 (mod 8). - K.V.Iyer, Apr 27 2009

References

  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 248.

Crossrefs

Cf. A004776 (complement), A007521 (primes).

Programs

Formula

From R. J. Mathar, Mar 14 2011: (Start)
a(n) = 8*n - 3.
G.f.: x*(5+3*x)/(x-1)^2. (End)
a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, May 28 2011
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: exp(x)*(8*x - 3) + 3.
a(n) = A113770(n)/2 = A016813(2*n-1). (End)
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