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 A037270 a(n) = n^2*(n^2 + 1)/2. 37
 0, 1, 10, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10440, 14365, 19306, 25425, 32896, 41905, 52650, 65341, 80200, 97461, 117370, 140185, 166176, 195625, 228826, 266085, 307720, 354061, 405450 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Sum of first n^2 positive integers. Start from xanthene and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002; Amarnath Murthy, Aug 01 2002 Sum of the next n multiples of n. - Amarnath Murthy, Aug 01 2002 The sum of the terms in an n X n spiral. These are also triangular numbers. - William A. Tedeschi, Feb 27 2008 Hypotenuse of Pythagorean triangles with smallest side a cube: A000578(n)^2 + A083374(n)^2 = a(n)^2. - Martin Renner, Nov 12 2011 For n>1, triangular numbers that can be represented as a sum of a square and a triangular number. For example, a(2)=10=4+6=9+1. - Ivan N. Ianakiev, Apr 24 2012 A037270 can be constructed in the following manner: Take A000217 and for every n not in A000290 delete the corresponding A000217(n). - Ivan N. Ianakiev, Apr 26 2012 Starting at a(1)=1 simply take 1*1=1, a(2)= 2*(2+3)=10, a(3)= 3*(4+5+6)=45, a(4)=4*(7+8+9+10) and so on. - J. M. Bergot, May 01 2015 Observation: The digital roots of the terms repeat in the sequence 1, 1, 9; e.g., the digital roots of 1, 10, 45, 136, 325, and 666 are 1, 1, 9, 1, 1, and 9. Verified for the first 10000 terms. - Rob Barton, Mar 28 2018 The above observation is easily explained and proved given that the digital root of a positive number equals the number modulo 9, and a(n + 9k) == a(n) (mod 9). - M. F. Hasler, Apr 05 2018 Number of unoriented rows of length 4 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=10, there are 4 achiral (AAAA, ABBA, BAAB, BBBB) and 6 chiral pairs (AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, BABB-BBAB). - Robert A. Russell, Nov 14 2018 REFERENCES C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 5. Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 55. T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002. T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445. R. A. Wilson, Cosmic Trigger, epilogue of S.-P. Sirag. LINKS T. D. Noe, Table of n, a(n) for n = 0..1000 J. D. Bell, A translation of Leonhard Euler's "De Quadratis Magicis", E795, arXiv:math/0408230 [math.CO], 2004-2005. N. G de Bruijn, Some classes of integer-valued functions, Nederl. Akad. Wetensch. Proc, Ser. A, 58 (1955), 363-367. See page 363. Th. Gruner, A. Kerber, R. Laue, M. Meringer, Mathematics for Combinatorial Chemistry, In: F. Keil, W. Mackens, H. Voß and J. Wether, Scientific Computing in Chemical Engineering II, Springer, 1999, 74-81. Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5, 1). FORMULA a(n) = a(n-1) + n^3 + (n-1)^3. a(n) = A000537(n)+A000537(n-1), i.e., square of sum of first n integers plus square of sum of first n-1 integers. - Henry Bottomley, Oct 15 2001 a(n) = Sum_{k=0..n^2} k. - William A. Tedeschi, Feb 27 2008 a(n) = (1/8)*sinh(2*arcsinh(n)). - Artur Jasinski, Feb 10 2010 G.f.: x*(1+x)*(1+4*x+x^2)/(1-x)^5. - Colin Barker, Mar 22 2012 a(n) = a(n-1) + A005898(n-1). - Ivan N. Ianakiev, May 13 2012 a(n) = 2 * A000217(n-1) * A000217(n) + A000290(n). - Ivan N. Ianakiev, May 26 2012 a(n) = A000217(n^2). - J. M. Bergot, Jun 07 2012 a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5) n>4, a(0)=0, a(1)=1, a(2)=10, a(3)=45, a(4)=136. - Yosu Yurramendi, Sep 02 2013 For n>0, a(n) = A000217(n)^2 + A000217(n-1)^2. - Richard R. Forberg, Dec 25 2013 a(n) = T(T(n)) + T(T(n-1)) + T(T(n)-1) + T(T(n-1)-1), where T(n) = A000217(n). - Charlie Marion, Sep 10 2016 a(n) = t(n-3)*t(n)+t(n-1)*t(n+2), with t(n)=A000217(n). - J. M. Bergot, Apr 07 2018 From Robert A. Russell, Nov 14 2018: (Start) a(n) = (A000583(n) + A000290(n)) / 2 = (n^4 + n^2) / 2. a(n) = A000583(n) - A083374(n) = A083374(n) + A000290(n). G.f.: (Sum_{j=1..4} S2(4,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..2} S2(2,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277. G.f.: Sum_{k=1..4} A145882(4,k) * x^k / (1-x)^5. E.g.f.: (Sum_{k=1..4} S2(4,k)*x^k + Sum_{k=1..2} S2(2,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277. For n>4, a(n) = Sum_{j=1..5} -binomial(j-6,j) * a(n-j). (End) a(n) = n*A006003(n). - Kritsada Moomuang, Dec 16 2018 For n > 0, a(n) = Sum_{k=1..n} A317617(n,k). - Stefano Spezia, Jan 10 2019 Sum_{n>=1} 1/a(n) = 1 + Pi^2/3 - Pi*coth(Pi) = 1.13652003875929052467672874379... - Vaclav Kotesovec, Jan 21 2019 MAPLE seq(n^2*(n^2+1)/2, n=0..30); # Muniru A Asiru, Mar 28 2018 MATHEMATICA Table[ n^2*((n^2 + 1)/2), {n, 0, 30} ] Table[(1/8) Round[N[Sinh[2 ArcSinh[n]]^2, 100]], {n, 0, 30}] (* Artur Jasinski, Feb 10 2010 *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 10, 45, 136}, 30] (* Harvey P. Dale, Aug 03 2014 *) PROG (PARI) a(n)=binomial(n^2+1, 2) \\ Charles R Greathouse IV, Apr 25 2012 (GAP) a:=List([0..30], n->n^2*(n^2+1)/2); # Muniru A Asiru, Mar 28 2018 (Python) for n in range(0, 30): print(n**2*(n**2+1)/2, end=', ') # Stefano Spezia, Jan 10 2019 (MAGMA) [n^2*(n^2 + 1)/2: n in [0..30]] // Stefano Spezia, Jan 15 2019 CROSSREFS Cf. A000217, A236770 (see crossrefs). Row 4 of A277504. Cf. A000583 (oriented), A083374 (chiral), A000290 (achiral). Cf. A317617. Sequence in context: A211032 A179095 A213188 * A027800 A005714 A175705 Adjacent sequences:  A037267 A037268 A037269 * A037271 A037272 A037273 KEYWORD nonn,easy,nice AUTHOR Aaron Gulliver (gulliver(AT)elec.canterbury.ac.nz) STATUS approved

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Last modified February 20 18:39 EST 2019. Contains 320345 sequences. (Running on oeis4.)