%I M4733 N2025 #173 Oct 04 2024 11:23:45
%S 0,1,10,91,820,7381,66430,597871,5380840,48427561,435848050,
%T 3922632451,35303692060,317733228541,2859599056870,25736391511831,
%U 231627523606480,2084647712458321,18761829412124890,168856464709124011,1519708182382116100,13677373641439044901,123096362772951404110
%N a(n) = (9^n - 1)/8.
%C From _David W. Wilson_: Numbers triangular, differences square.
%C To be precise, the differences are the squares of the powers of three with positive indices. Hence a(n+1) - a(n) = (A000244(n+1))^2 = A001019(n+1). [Added by _Ant King_, Jan 05 2011]
%C Partial sums of A001019. This is m-th triangular number, where m is partial sums of A000244. a(n) = A000217(A003462(n)). - _Lekraj Beedassy_, May 25 2004
%C With offset 0, binomial transform of A003951. - _Philippe Deléham_, Jul 22 2005
%C Numbers in base 9: 1, 11, 111, 1111, 11111, 111111, 1111111, etc. - _Zerinvary Lajos_, Apr 26 2009
%C Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=9, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - _Milan Janjic_, Feb 21 2010
%C Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 2, a(n-1) = (-1)^n*charpoly(A,1). - _Milan Janjic_, Feb 21 2010
%C From _Hieronymus Fischer_, Jan 30 2013: (Start)
%C Least index k such that A052382(k) >= 10^(n-1), for n > 0.
%C Also index k such that A052382(k) = (10^n-1)/9, n > 0.
%C A052382(a(n)) is the least zerofree number with n digits, for n > 0.
%C For n > 1: A052382(a(n)-1) is the greatest zerofree number with n-1 digits. (End)
%C For n > 0, 4*a(n) is the total number of holes in a certain triangle fractal (start with 9 triangles, 4 holes) after n iterations. See illustration in links. - _Kival Ngaokrajang_, Feb 21 2015
%C For n > 0, a(n) is the sum of the numerators and denominators of the reduced fractions 0 < (b/3^(n-1)) < 1 plus 1. Example for n=3 gives fractions 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, and 8/9 plus 1 has sum of numerators and denominators +1 = a(3) = 91. - _J. M. Bergot_, Jul 11 2015
%C Except for 0 and 1, all terms are Brazilian repunits numbers in base 9, so belong to A125134. All these terms are composite because a(n) is the ((3^n - 1)/2)-th triangular number. - _Bernard Schott_, Apr 23 2017
%C These are also the second steps after the junctions of the Collatz trajectories of 2^(2k-1)-1 and 2^2k-1. - _David Rabahy_, Nov 01 2017
%D A. Fletcher, J. C. P. Miller, L. Rosenhead, and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
%D J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D T. N. Thiele, Interpolationsrechnung. Teubner, Leipzig, 1909, p. 36.
%H Vincenzo Librandi, <a href="/A002452/b002452.txt">Table of n, a(n) for n = 0..300</a>
%H Carlos M. da Fonseca and Anthony G. Shannon, <a href="https://doi.org/10.7546/nntdm.2024.30.3.491-498">A formal operator involving Fermatian numbers</a>, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
%H Kival Ngaokrajang, <a href="/A002452/a002452.pdf">Illustration of initial terms</a>
%H Vladimir Pletser, <a href="http://arxiv.org/abs/1409.7969">Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers</a>, arXiv:1409.7969 [math.NT], 2014.
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H A. G. Shannon, <a href="/A003462/a003462_1.pdf">Letter to N. J. A. Sloane, Dec 06 1974</a>
%H M. Ward, <a href="http://dx.doi.org/10.1090/S0002-9904-1936-06435-9">Note on divisibility sequences</a>, Bull. Amer. Math. Soc., 42 (1936), 843-845.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Repunit.html">Repunit</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-9).
%F From _Philippe Deléham_, Mar 13 2004: (Start)
%F a(n) = 9*a(n-1) + 1; a(1) = 1.
%F G.f.: x / ((1-x)*(1-9*x)). (End)
%F a(n) = 10*a(n-1) - 9*a(n-2). - _Ant King_, Jan 05 2011
%F a(n) = floor(A000217(3^n)/4) - A033113(n-1). - _Arkadiusz Wesolowski_, Feb 14 2012
%F Sum_{n>0} a(n)*(-1)^(n+1)*x^(2*n+1)/(2*n+1)! = (1/6)*sin(x)^3. - _Vladimir Kruchinin_, Sep 30 2012
%F a(n) = A011540(A217094(n-1)) - A217094(n-1) + 2, n > 0. - _Hieronymus Fischer_, Jan 30 2013
%F a(n) = 10^(n-1) + 2 - A217094(n-1). - _Hieronymus Fischer_, Jan 30 2013
%F a(n) = det(|v(i+2,j+1)|, 1 <= i,j <= n-1), where v(n,k) are central factorial numbers of the first kind with odd indices (A008956) and n > 0. - _Mircea Merca_, Apr 06 2013
%F a(n) = Sum_{k=0..n-1} 9^k. - _Doug Bell_, May 26 2017
%F E.g.f.: exp(5*x)*sinh(4*x)/4. - _Stefano Spezia_, Mar 11 2023
%e a(4) = (9^4 - 1)/8 = 820 = 1111_9 = (1/2) * 40 * 41 is the ((3^4 - 1)/2)-th = 40th triangular number. - _Bernard Schott_, Apr 23 2017
%p A002452 := 1/(9*z-1)/(z-1); # _Simon Plouffe_ in his 1992 dissertation
%t (9^# & /@ Range[0, 18] // Accumulate) (* _Ant King_, Jan 06 2011 *)
%t LinearRecurrence[{10,-9},{0,1},30] (* _Harvey P. Dale_, Sep 23 2018 *)
%o (Magma) [(9^n - 1)/8 : n in [0..25]]; // _Vincenzo Librandi_, Jun 01 2011
%o (PARI) a(n)=9^n>>3 \\ _Charles R Greathouse IV_, Jul 25 2011
%o (Maxima) A002452(n):=floor((9^n-1)/8)$
%o makelist(A002452(n),n,0,30); /* _Martin Ettl_, Nov 05 2012 */
%Y Right-hand column 1 in triangle A008958.
%Y Cf. A217094, A011540, A052382, A125857.
%Y Cf. A125134, A000217.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
%E More terms from Pab Ter (pabrlos(AT)yahoo.com), May 08 2004
%E Offset changed from 1 to 0 and added 0 by _Vincenzo Librandi_, Jun 01 2011