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 10 results.

A187869 a(n) is the concatenation of the numbers appearing in row n of the lexicographically earliest prime pyramid (A051237).

Original entry on oeis.org

1, 12, 123, 1234, 14325, 143256, 1432567, 12347658, 123476589, 12347658910, 1234710985611, 123471098561112, 12347651211891013, 1234761310981112514, 123476131098111251415, 12347651211815149101316, 1234765121189101316151417, 123476589101316151417121118
Offset: 1

Views

Author

N. J. A. Sloane, Mar 14 2011

Keywords

Crossrefs

Cf. A051237.

A055265 a(n) is the smallest positive integer not already in the sequence such that a(n)+a(n-1) is prime, starting with a(1)=1.

Original entry on oeis.org

1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 16, 15, 14, 17, 12, 11, 18, 19, 22, 21, 20, 23, 24, 29, 30, 31, 28, 25, 34, 27, 26, 33, 38, 35, 32, 39, 40, 43, 36, 37, 42, 41, 48, 49, 52, 45, 44, 53, 50, 47, 54, 55, 46, 51, 56, 57, 70, 61, 66, 65, 62, 69, 58, 73, 64, 63, 68, 59, 72, 67, 60
Offset: 1

Views

Author

Henry Bottomley, May 09 2000

Keywords

Comments

The sequence is well-defined (the terms must alternate in parity, and by Dirichlet's theorem a(n+1) always exists). - N. J. A. Sloane, Mar 07 2017
Does every positive integer eventually occur? - Dmitry Kamenetsky, May 27 2009. Reply from Robert G. Wilson v, May 27 2009: The answer is almost certainly yes, on probabilistic grounds.
It appears that this is the limit of the rows of A051237. That those rows do approach a limit seems certain, and given that that limit exists, that this sequence is the limit seems even more likely, but no proof is known for either conjecture. - Robert G. Wilson v, Mar 11 2011, edited by Franklin T. Adams-Watters, Mar 17 2011
The sequence is also a particular case of "among the pairwise sums of any M consecutive terms, N are prime", with M = 2, N = 1. For other M, N see A055266 & A253074 (M = 2, N = 0), A329333, A329405 - A329416, A329449 - A329456, A329563 - A329581, and the OEIS Wiki page. - M. F. Hasler, Feb 11 2020

Examples

			a(5) = 7 because 1, 2, 3 and 4 have already been used and neither 4 + 5 = 9 nor 4 + 6 = 10 are prime while 4 + 7 = 11 is prime.
		

Crossrefs

Inverse permutation: A117922; fixed points: A117925; A117923=a(a(n)). - Reinhard Zumkeller, Apr 03 2006
Cf. A086527 (the primes a(n)+a(n-1)).
Cf. A070942 (n's such that a(1..n) is a permutation of (1..n)). - Zak Seidov, Oct 19 2011
See also A076990, A243625.
See A282695 for deviation from identity sequence.
A073659 is a version where the partial sums must be primes.

Programs

  • Haskell
    import Data.List (delete)
    a055265 n = a055265_list !! (n-1)
    a055265_list = 1 : f 1 [2..] where
       f x vs = g vs where
         g (w:ws) = if a010051 (x + w) == 1
                       then w : f w (delete w vs) else g ws
    -- Reinhard Zumkeller, Feb 14 2013
    
  • Maple
    A055265 := proc(n)
        local a,i,known ;
        option remember;
        if n =1 then
            1;
        else
            for a from 1 do
                known := false;
                for i from 1 to n-1 do
                    if procname(i) = a then
                        known := true;
                        break;
                    end if;
                end do:
                if not known and isprime(procname(n-1)+a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A055265(n),n=1..100) ; # R. J. Mathar, Feb 25 2017
  • Mathematica
    f[s_List] := Block[{k = 1, a = s[[ -1]]}, While[ MemberQ[s, k] || ! PrimeQ[a + k], k++ ]; Append[s, k]]; Nest[f, {1}, 71] (* Robert G. Wilson v, May 27 2009 *)
    q=2000; a={1}; z=Range[2,2*q]; While[Length[z]>q-1, k=1; While[!PrimeQ[z[[k]]+Last[a]], k++]; AppendTo[a,z[[k]]]; z=Delete[z,k]]; Print[a] (*200 times faster*) (* Vladimir Joseph Stephan Orlovsky, May 03 2011 *)
  • PARI
    v=[1];n=1;while(n<50,if(isprime(v[#v]+n)&&!vecsearch(vecsort(v),n), v=concat(v,n);n=0);n++);v \\ Derek Orr, Jun 01 2015
    
  • PARI
    U=-a=1; vector(100,k, k=valuation(1+U+=1<M. F. Hasler, Feb 11 2020

Formula

a(2n-1) = A128280(2n-1) - 1, a(2n) = A128280(2n) + 1, for all n >= 1. - M. F. Hasler, Feb 11 2020

Extensions

Corrected by Hans Havermann, Sep 24 2002

A010551 Multiply successively by 1,1,2,2,3,3,4,4,..., n >= 1, a(0) = 1.

Original entry on oeis.org

1, 1, 1, 2, 4, 12, 36, 144, 576, 2880, 14400, 86400, 518400, 3628800, 25401600, 203212800, 1625702400, 14631321600, 131681894400, 1316818944000, 13168189440000, 144850083840000, 1593350922240000, 19120211066880000, 229442532802560000, 2982752926433280000
Offset: 0

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Author

Keywords

Comments

From Emeric Deutsch, Dec 14 2008: (Start)
Number of permutations of {1,2,...,n-1} having a single run of odd entries. Example: a(5)=12 because we have 1324,1342,3124,3142,2134,4132,2314,4312, 2413, 4213, 2431 and 4231.
a(n) = A152666(n-1,1). (End)
a(n+1) gives the permanent of the n X n matrix whose (i,j)-element is i+j-1 modulo 2. - John W. Layman, Jan 03 2011
From Daniel Forgues, May 20 2011: (Start)
a(0) = 1 since it is the empty product.
A010551(n-2), n >= 2, equal to (ceiling((n-2)/2))! * (floor((n-2)/2))!, gives the number of arrangements of n-2 entries from 2 to n-1, starting with an even entry and where the parity of adjacent entries alternates. This is the number of arrangements to investigate for row n of a prime pyramid (A051237). (End)
Partial products of A008619. - Reinhard Zumkeller, Apr 02 2012
Also size of the equivalence class of S_n containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> acb where a < b < c, cf. A210667 (equivalently under such transformations of the form abc <--> bac where a < b < c.) - Tom Roby, May 15 2012
Row sums of A246117. - Peter Bala, Aug 15 2014
a(n) is the number of parity-alternating permutations of size n. A permutation is parity-alternating if it sends even integers to even, and odd to odd. - Per W. Alexandersson, Jun 06 2022
n divides a(n) if and only if n is not prime. Since a(n) = floor(n/2)!*floor((n+1)/2)!, if n is prime then n is not a factor of a(n). All the prime factors of a(n) are in fact less than or equal to (n+1)/2. If n is composite, then it's possible to write it as p*q with p and q less than or equal to n/2. So p and q are factors of a(n). - Davide Oliveri, Apr 01 2023
Number of permutations of {1, 2, ..., n-1} where each entry is not greater than twice the previous entry. - Dewangga Putra Sheradhien, Jul 13 2024

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 4*x^4 + 12*x^5 + 36*x^6 + 144*x^7 + 576*x^8 + ...
For n = 7, a(n) = 1*1*2*2*3*3*4 (7 factors), which is 144. - _Michael B. Porter_, Jul 03 2016
		

Crossrefs

Column k=2 of A275062.

Programs

  • Haskell
    a010551 n = a010551_list !! n
    a010551_list = scanl (*) 1 a008619_list
    -- Reinhard Zumkeller, Apr 02 2012
    
  • Magma
    [Factorial(n div 2)*Factorial((n+1) div 2): n in [0..25]]; // Vincenzo Librandi Jan 17 2018
    
  • Maple
    A010551 := proc(n)
        option remember;
        if n <= 1 then
            1
        else
            procname(n-1) *trunc( (n+1)/2 );
        fi;
    end:
  • Mathematica
    FoldList[ Times, 1, Flatten@ Array[ {#, #} &, 11]] (* Robert G. Wilson v, Jul 14 2010 *)
  • PARI
    {a(n)=local(X=x+x*O(x^n)); 1/polcoeff(besseli(0,2*X)+X*besseli(1,2*X),n,x)} \\ Paul D. Hanna, Apr 07 2005
    
  • PARI
    A010551(n)=(n\2)!*((n+1)\2)! \\ Michael Somos, Dec 29 2012, edited by M. F. Hasler, Nov 26 2017
    
  • Python
    def O(f):
        c = 1
        while len(f) > 1:
            f.sort()
            m = abs(f[0] - f[1])
            c *= m
            f[0] = m
            f.pop(1)
        return c
    a = lambda n: O(list(range(1, n+1)))
    print([a(n) for n in range(0, 26)]) # Darío Clavijo, Aug 24 2024

Formula

a(n) = floor(n/2)!*floor((n+1)/2)! is the number of permutations p of {1, 2, 3, ..., n} such that for every i, i and p(i) have the same parity, i.e., p(i) - i is even. - Avi Peretz (njk(AT)netvision.net.il), Feb 22 2001
a(n) = n!/binomial(n, floor(n/2)). - Paul Barry, Sep 12 2004
G.f.: Sum_{n>=0} x^n/a(n) = besseli(0, 2*x) + x*besseli(1, 2*x). - Paul D. Hanna, Apr 07 2005
E.g.f.: 1/(1-x/2) + (1/2)/(1-x/2)*arccos(1-x^2/2)/sqrt(1-x^2/4). - Paul D. Hanna, Aug 28 2005
G.f.: G(0) where G(k) = 1 + (k+1)*x/(1 - x*(k+1)/(x*(k+1) + 1/G(k+1) )); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 28 2012
D-finite with recurrence: 4*a(n) - 2*a(n-1) - n*(n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
a(n) = a(n-1) * (a(n-2) + a(n-3)) / a(n-3) for all n >= 3. - Michael Somos, Dec 29 2012
G.f.: 1 + x + x^2*(1 + x*(G(0) - 1)/(x-1)) where G(k) = 1 - (k+2)/(1-x/(x - 1/(1 - (k+2)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - (k+1)/(1-x/(x - 1/(1 - (k+1)/(1-x/(x - 1/G(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: 1 + x*G(0), where G(k) = 1 + x*(k+1)/(1 - x*(k+2)/(x*(k+2) + 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 08 2013
G.f.: Q(0), where Q(k) = 1 + x*(k+1)/(1 - x*(k+1)/(x*(k+1) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 08 2013
Sum_{n >= 1} 1/a(n) = A130820. - Peter Bala, Jul 02 2016
a(n) ~ sqrt(Pi*n) * n! / 2^(n + 1/2). - Vaclav Kotesovec, Oct 02 2018
Sum_{n>=0} (-1)^n/a(n) = A229020. - Amiram Eldar, Apr 12 2021

A051239 Number of ways to arrange integers 1 through n so that the sum of each adjacent pair is prime, not counting reversals.

Original entry on oeis.org

1, 1, 1, 4, 2, 8, 12, 30, 70, 664, 1072, 8768, 11648, 37108, 95772, 1059816, 2047488, 12111712, 22802028, 120779959, 337801784, 4361743860, 11425028900, 142573286216, 429917269469, 4138239848098, 16171519847028, 214845911686065
Offset: 1

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Author

Keywords

Comments

Similar to A036440, but not requiring 1 to be first and n to be last.

Examples

			a(5)=2 since 1,4,3,2,5 and 3,4,1,2,5 have the property that the sum of each adjacent pair is prime.
		

Crossrefs

Formula

For n>1, a(n) = A103839(n)/2.

Extensions

More terms from Max Alekseyev, Jan 04 2008
a(25)-a(28) from Giovanni Resta, Apr 01 2014

A036440 Number of ways of arranging row n of the prime pyramid.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 4, 7, 24, 80, 216, 648, 1304, 3392, 13808, 59448, 155464, 480728, 1588162, 5626309, 28279112, 157469880, 842498189, 4998554801, 28466978744, 166572523589, 1020487969891, 5753385660978
Offset: 1

Views

Author

Keywords

Comments

Number of ways to arrange numbers from 1 to n in a row, starting with 1 and ending with n, such that the sum of every two adjacent numbers is prime.
From Daniel Forgues, May 18 2011: (Start)
Since the sum of any two adjacent entries is at least 3, the sum is thus an odd prime, which implies that any two consecutive entries have opposite parity. Since the first and last entries of row n are fixed to 1 and n, we have to find n-2 entries, where ceiling((n-2)/2) are even and floor((n-2)/2) are odd, so for row n the number of arrangements to investigate is
(ceiling((n-2)/2))! * (floor((n-2)/2))! (Cf. A010551(n-2), n >= 2.)
Prime pyramids are also (more fittingly?) called prime triangles. (End)

Examples

			a(8)=4 because of the 720 permutations P of {1,2,3,4,5,6,7,8} with first element 1 and last element 8, these four satisfy the "prime pyramid" condition that P[i] + P[i+1] be prime for i=1..7:
  1 2 3 4 7 6 5 8; (lexicographically earliest row 8)
  1 2 5 6 7 4 3 8;
  1 4 7 6 5 2 3 8;
  1 6 7 4 3 2 5 8.
For row 8, there are 6! = 720 permutations of {2,3,4,5,6}, but if we take into account that the parity of all entries of row n must alternate, we only have to consider ceiling((8-2)/2)! * floor((8-2)/2)! = 36 cases.
		

References

  • R. K. Guy, Unsolved Problems Number Theory, C1.

Crossrefs

Cf. A051237 for lexicographically earliest prime pyramid.

Programs

  • Mathematica
    prsQ[n_]:=AllTrue[Total/@Partition[n,2,1],PrimeQ]; Table[Count[Join[{1},#,{n}]&/@ Permutations[ Range[2,n-1]],?prsQ],{n,12}] (* _Harvey P. Dale, May 07 2023 *)

Extensions

More terms from Jud McCranie
a(25)-a(27) from Max Alekseyev, Jan 05 2008
a(28)-a(29) from Giovanni Resta, Apr 01 2014

A124886 3-almost prime triangle, read by rows.

Original entry on oeis.org

1, 1, 7, 1, 11, 9, 1, 17, 3, 5, 1, 19, 8, 4, 14, 1, 26, 2, 6, 12, 15, 1, 27, 18, 10, 20, 30, 22, 1, 29, 13, 31, 21, 23, 40, 28, 1, 41, 25, 38, 32, 34, 16, 36, 39, 1, 43, 33, 35, 57, 42, 24, 44, 48, 50
Offset: 1

Views

Author

Jonathan Vos Post, Nov 12 2006

Keywords

Comments

This is to 3-almost primes (A014612) as A124883 is to semiprimes (A001358). The n-th row is of length n. Each value is the smallest previously unused natural number such that every pair of adjacent values in the triangle is 3-almost prime (A014612). Consider row 2. Starting with T(1,2) = 1, the least integer we can add to 1 and get a 3-almost prime is 7, since 1 + 8 = 8 = 2^3 is 3-almost prime. Consider row 3. Starting with T(1,3) = 1, the least integer we can add to 1 and get a 3-almost prime is 7, but we've already used that. The least unused integer that works is 11, since 1 + 11 = 12 = 2^2 * 3 is 3-almost prime. If we cross out ones from the triangle read by rows, what remains is a permutation of the natural number greater than 1. That is, every nonnegative integer appears in the triangle. The second column T(n,2) is monotone increasing.

Examples

			Triangle begins:
  1
  1  7
  1 11  9
  1 17  3  5
  1 19  8  4 14
  1 26  2  6 12 15
  1 27 18 10 20 30 22
  1 29 13 31 21 23 40 28
  1 41 25 38 32 34 16 36 39
  1 43 33 35 57 42 24 44 48 50
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, p. 106, 1994.
  • M. J. Kenney, "Student Math Notes." NCTM News Bulletin. Nov. 1986.

Crossrefs

Formula

T(n,1) = 1 for all natural numbers n. For n>1 and 1T(n,i) for iT(r,s) for rA014612}.

A124883 Semiprime triangle, read by rows.

Original entry on oeis.org

1, 1, 3, 1, 5, 4, 1, 8, 2, 7, 1, 9, 6, 15, 10, 1, 13, 12, 14, 11, 22, 1, 20, 18, 16, 17, 21, 25, 1, 24, 27, 19, 30, 28, 23, 26, 1, 32, 33, 29, 36, 38, 31, 34, 35, 1, 37, 40, 42, 43, 39, 46, 41, 44, 47, 1, 45, 48, 58, 53, 62, 49, 57, 54, 52, 59
Offset: 1

Views

Author

Jonathan Vos Post, Nov 11 2006

Keywords

Examples

			The n-th row is of length n. Each value is the smallest previously unused natural number such that the sum of every pair of adjacent values in the triangle is a semiprime (A001358).
Consider row 2. Starting with T(1,2) = 1, the least integer we can add to 1 and get a semiprime is 3, since 1 + 3 = 4 = 2^2 is semiprime. Consider row 3. Starting with T(1,3) = 1, the least integer we can add to 1 and get a semiprime is 1, but we've already used that. The next is 3, but we've used that. The least unused integer that works is 5, since 1 + 5 = 6 = 2 * 3 is semiprime. If we cross out ones from the triangle read by rows, what remains is a permutation of the natural number greater than 1. That is, every nonnegative integer appears in the triangle. The second column T(n,2) is monotone increasing.
Triangle begins:
  1;
  1,  3;
  1,  5,  4;
  1,  8,  2,  7;
  1,  9,  6, 15, 10;
  1, 13, 12, 14, 11, 22;
  1, 20, 18, 16, 17, 21, 25;
  1, 24, 27, 19, 30, 28, 23, 26;
  1, 32, 33, 29, 36, 38, 31, 34, 35;
  1, 37, 40, 42, 43, 39, 46, 41, 44, 47;
  1, 45, 48, 58, 53, 62, 49, 57, 54, 52, 59;
  ...
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, p. 106, 1994.
  • M. J. Kenney, "Student Math Notes." NCTM News Bulletin. Nov. 1986.

Crossrefs

Formula

T(n,1) = 1 for all natural numbers n. For n>1 and 1T(n,i) for iT(r,s) for rA001358}.

Extensions

Terms corrected by Alois P. Heinz, Apr 08 2025

A134292 Triangle in which row n is the lexicographically earliest solution to the prime circle problem for 2n.

Original entry on oeis.org

1, 2, 1, 2, 3, 4, 1, 4, 3, 2, 5, 6, 1, 2, 3, 8, 5, 6, 7, 4, 1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 1, 2, 3, 4, 7, 6, 5, 12, 11, 8, 9, 10, 1, 2, 3, 4, 7, 6, 13, 10, 9, 14, 5, 8, 11, 12, 1, 2, 3, 4, 7, 6, 5, 12, 11, 8, 9, 14, 15, 16, 13, 10, 1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 16, 15, 14, 17, 12, 11, 18, 1, 2, 3
Offset: 1

Views

Author

T. D. Noe, Oct 16 2007

Keywords

Comments

In the prime circle problem we seek to arrange the numbers 1 to 2n around a circle so that the sum of each pair of adjacent numbers is prime. To display the solution, we unroll the circle starting at 1.

Examples

			Triangle begins:
1, 2;
1, 2, 3, 4;
1, 4, 3, 2, 5, 6;
1, 2, 3, 8, 5, 6, 7, 4;
...
		

References

  • R. K. Guy, Unsolved Problems Number Theory, Section C1.

Crossrefs

Cf. A051252 (number of solutions for each n), A051237 (prime pyramid).

A187871 a(1)=1; for n > 1, a(n) is the smallest number that is formed by arranging the decimal numbers "1", "2", ..., "n" in some order so that the sum of every pair of adjacent numbers "i" "j" is prime.

Original entry on oeis.org

1, 12, 123, 1234, 14325, 143256, 1234765, 12347658, 123476589, 10123476589, 1103211658947, 101121123476589, 11013432111276589, 1011211231458941367, 110134151432111276589, 10112112151431613476589, 1101316151417121123476589, 101121118131615141723476589
Offset: 1

Views

Author

N. J. A. Sloane, Mar 14 2011

Keywords

Comments

A051237 and A187869 are the sequence that result if in addition we require that the number begins with "1" and ends with "n".

Crossrefs

Programs

  • PARI
    See Links section.

Extensions

a(6) to a(10) from Claudio Meller, Mar 14 2011
a(10) corrected by Nathaniel Johnston, Mar 15 2011
More terms from Rémy Sigrist, Nov 28 2021

A124939 Prime tetrahedron, read by rows.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 4, 1, 6, 5, 1, 1, 10, 1, 12, 7, 1, 16, 3, 8, 1, 1, 18, 1, 22, 9, 1, 28, 13, 24, 1, 30, 11, 20, 17, 1, 1, 36, 1, 40, 19, 1, 42, 25, 34, 1, 46, 15, 14, 23, 1, 52, 21, 26, 27, 32, 1, 1, 58, 1, 60, 29, 1, 66, 31, 48, 1, 70, 33, 38, 35, 1, 72, 37, 64, 39, 44, 1, 78, 49, 54, 43
Offset: 1

Views

Author

Jonathan Vos Post, Nov 13 2006

Keywords

Comments

Each triangular layer of the unique tetrahedron begins with 1, never uses any value other than 1 which has occurred already on this or earlier levels, always uses the least available integer such that the sum of each two consecutive entries is a prime. The number of values of the n-th level is the n-th triangular number A000217(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n. The number of values through the n-th level is the n-th tetrahedral number A000292(n) = C(n+2,3) = n(n+1)(n+2)/6.

Examples

			Tetrahedron begins
=================
1
=================
1
1..2
=================
1
1..4
1..6..5
=================
1
1.10
1.12..7
1.16..3..8
=================
1
1.18
1.22..9
1.28.13.24
1.30.11.20.17
=================
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, p. 106, 1994.
  • Kenney, M. J. "Student Math Notes." NCTM News Bulletin. Nov. 1986.

Crossrefs

Cf. A000040, A000217, A000292, A036440 Number of ways of arranging row n of the Prime Pyramid, A051239, A051237 Lexicographically earliest Prime Pyramid, read by rows.

Programs

  • Maple
    srch := proc(a) local res ; res := 2 ; while true do if isprime(res+op(-1,a)) and not ( res in a ) then RETURN(res) ; fi ; res := res+1 ; od ; end: a := [] ; for lvl from 1 to 10 do for row from 1 to lvl do for col from 1 to row do if col = 1 then anxt := 1 ; else anxt := srch(a) ; fi ; printf("%d,",anxt) ; a := [op(a), anxt] ; od ; od ; od ; # R. J. Mathar, Jan 13 2007

Formula

a(n) flattens the 3-D table so that level 1 (the apex, with only the value 1) occurs first, then level 2 (with values 1, 1, 2), then level 3 ... and for each level, reads that triangle by rows.

Extensions

Corrected and extended by R. J. Mathar, Jan 13 2007
Showing 1-10 of 10 results.